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

Question

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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.

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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.

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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.

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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.

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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 .

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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

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