The line integral V A B along the line that joins the points A and B .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

PROBLEM 2.50 1.8 m The concrete post (E-25 GPa and a = 9.9 x 10°/°C) is reinforced with six steel bars, each of 22-mm diameter (E, = 200 GPa and a, = 11.7 x 10°/°C). Determine the normal stresses induced in the steel and in the concrete by a temperature rise of 35°C. 6c " 0.391 MPa 240 mm 240 mm 6₁ = -9.47 MPa

For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to a(t) = a(0) exp(-4) (15.10) where σ(t) and o(0) represent the time-dependent and initial (i.e., time = 0) stresses, respectively, and t and T denote elapsed time and the relaxation time, respectively; T is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.5; the stress necessary to maintain this constant strain was measured as a function of time. Determine E (10) for this material if the initial stress level was 3.5 MPa (500 psi), which dropped to 0.5 MPa (70 psi) after 30 s.

For the flows in Examples 11.1 and 11.2, calculate the magnitudes of the Δ V2 / 2 terms omitted in B.E., and compare these with the magnitude of the ℱ terms.

Answer 2

Question

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Answer 6

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

Answer 7

Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral VAB along the line that joins the points A and B.

Answer 8

(i)

Expert Solution

Answer to Problem 2.8MA

The line integral VAB along the line that joins the points A and B is –1039.

Answer 9

(i)

Expert Solution

Answer 10

(i)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment.

Obtain the parameterization of the line as follows.

l→(t)=(1−t)(3ax+7ay+az)+t(8ax+9ay+2az)=[3(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+2t]az=(3−3t+8t)ax+(7−7t+9t)ay+(1−t+2t)az=(3+5t)ax+(7+2t)ay+(1+t)az

The vector along the line segment becomes

E=(4(3+5t)(7+2t)+(1+t))ax+2(3+5t)2ay+(3+5t)az=(4(21+6t+35t+10t2)+(1+t))ax+2(9+30t+25t2)ay+(3+5t)az=(84+24t+140t+40t2+1+t)ax+18+60t+50t2ay+(3+5t)az=(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az

Obtain the derivative of the parameterization as dl=l→′(t)=[5ax+2ay+az]dt.

Perform the line integration as follows.

∫CE⋅dl=∫01[(85+165t+40t2)ax+(18+60t+50t2)ay+(3+5t)az]⋅[5ax+2ay+az]dt=∫01[5(85+165t+40t2)+2(18+60t+50t2)+(3+5t)]dt=∫015(85+165t+40t2)dt+∫012(18+60t+50t2)dt+∫01(3+5t)dt=5[85t+165t22+40t33]01+2[18t+60t22+50t33]01+[3t+5t22]01

On further simplification,

∫CE⋅dl=5[85+16512+4013]+2[18+6012+5013]+[3+512]=425+8252+2003+36+60+1003+3+52=524+415+100=1039

Since VAB=−∫CE⋅dl, the value of VAB becomes VAB=−1039

Therefore, the line integral VAB along the line that joins the points A and B is –1039.

Answer 13

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Answer 14

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

Answer 15

(ii)

Expert Solution

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Answer 16

(ii)

Expert Solution

Answer 17

(ii)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The vector is given as E=(4xy+z)ax+2x2ay+xaz and the points are A(3,7,1) to B(8,9,2) that forms the line segment with stopovers C and D are C(8,7,1) and D(8,9,1).

Obtain the parameterization of the line A to C as follows.

l1→(t)=(1−t)(3ax+7ay+az)+t(8ax+7ay+1az)=[3(1−t)+8t]ax+[7(1−t)+7t]ay+[1−t+t]az=(3−3t+8t)ax+(7−7t+7t)ay+(1)az=(3+5t)ax+7ay+az

Obtain the parameterization of the line C to D as follows.

l2→(t)=(1−t)(8ax+7ay+az)+t(8ax+9ay+az)=[8(1−t)+8t]ax+[7(1−t)+9t]ay+[1−t+t]az=(8−8t+8t)ax+(7−7t+9t)ay+(1)az=8ax+(7+2t)ay+az

Obtain the parameterization of the line D to B as follows.

l3→(t)=(1−t)(8ax+9ay+az)+t(8ax+9ay+2az)=[8(1−t)+8t]ax+[9(1−t)+9t]ay+[1−t+2t]az=(8−8t+8t)ax+(9−9t+9t)ay+(1+t)az=8ax+9ay+(1+t)az

The vector along the line segment l1→ becomes

E1=(4(3+5t)(7)+1)ax+2(3+5t)2ay+(3+5t)az=(84+140t+1)ax+2(9+30t+25t2)ay+(3+5t)az=(140t+85)ax+18+60t+50t2ay+(3+5t)az

The vector along the line segment l2→ becomes

E2=(4(8)(7+2t)+1)ax+2(8)2ay+(8)az=(224+64t+1)ax+2(64)ay+8az=(64t+225)ax+128ay+8az

The vector along the line segment l3→ becomes

E3=(4(8)(9)+1+t)ax+2(8)2ay+(8)az=(288+t+1)ax+2(64)ay+8az=(t+289)ax+128ay+8az

Obtain the derivative of the parameterizations as dl1=l1→′(t)=5axdt, dl2=l2→′(t)=2aydt and dl3=l3→′(t)=azdt.

Perform the line integration as follows.

∫CE1⋅dl1=∫01[(140t+85)ax+18+60t+50t2ay+(3+5t)az]⋅[5ax]dt=∫01[5(140t+85)]dt=5[140t22+85t]01=5[70+85]

On further simplification,

∫CE1⋅dl1=5[155]=775

Perform the line integration as follows.

∫CE2⋅dl2=∫01[(64t+225)ax+128ay+8az]⋅[2ay]dt=∫01[2(128)]dt=5[256t]01=5[256]

On further simplification, the integral becomes ∫CE2⋅dl2=1280.

Perform the line integration as follows.

∫CE3⋅dl3=∫01[(t+289)ax+128ay+8az]⋅[ay]dt=∫01[8]dt=[8t]01=8

Substitute the values of line integrals in the given expression and obtain the result.

−∫ABE⋅dl−∫CDE⋅dl−∫DBE⋅dl=−∫CE1⋅dl1−∫CE2⋅dl2−∫CE3⋅dl3=−775−1280−8=−2063

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1) is –2063.

Answer 20

Ch. MA - Prob. 1.1MA Ch. MA - Prob. 1.2MA Ch. MA - Prob. 1.3MA Ch. MA - Prob. 1.4MA Ch. MA - Prob. 1.5MA Ch. MA - Prob. 1.6MA Ch. MA - Prob. 1.7MA Ch. MA - Prob. 1.8MA Ch. MA - Prob. 1.9MA Ch. MA - Prob. 1.10MA

The line integral V A B along the line that joins the points A and B .

Concept explainers

Videos

To evaluate: The line integral VAB along the line that joins the points A and B.

Answer to Problem 2.8MA

Explanation of Solution

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are C(8,7,1) and D(8,9,1).

Answer to Problem 2.8MA

Explanation of Solution

Want to see more full solutions like this?

Chapter MA Solutions