N -2 -1.5 2 1.5 -1 -0.51 -0.5 -1 -1.5 -2 y 1.5 2 1.5 X 2 (x, y, z) = (cos z, sin z, z), − < z < Explanation: Step 1: y = x tan(z) with: x>0, x²+ y² = 1 −√ < ± < √ Step 2: sin(z) y = x cos(2) Step 3: substitute y: x²+x2. sin². cos2 Then we get: = 2 . (cos's+sin= 2) Hence we would get: x² = cos² z Step 4: Since x0, cos z> 0 We would get: x = COS z 1 sin(z) y = cos z. == sin z cos(2) Hence the parametric form is: (x, y, z) = (cos z, sin z, z), — √ <≤ z <
N -2 -1.5 2 1.5 -1 -0.51 -0.5 -1 -1.5 -2 y 1.5 2 1.5 X 2 (x, y, z) = (cos z, sin z, z), − < z < Explanation: Step 1: y = x tan(z) with: x>0, x²+ y² = 1 −√ < ± < √ Step 2: sin(z) y = x cos(2) Step 3: substitute y: x²+x2. sin². cos2 Then we get: = 2 . (cos's+sin= 2) Hence we would get: x² = cos² z Step 4: Since x0, cos z> 0 We would get: x = COS z 1 sin(z) y = cos z. == sin z cos(2) Hence the parametric form is: (x, y, z) = (cos z, sin z, z), — √ <≤ z <
N -2 -1.5 2 1.5 -1 -0.51 -0.5 -1 -1.5 -2 y 1.5 2 1.5 X 2 (x, y, z) = (cos z, sin z, z), − < z < Explanation: Step 1: y = x tan(z) with: x>0, x²+ y² = 1 −√ < ± < √ Step 2: sin(z) y = x cos(2) Step 3: substitute y: x²+x2. sin². cos2 Then we get: = 2 . (cos's+sin= 2) Hence we would get: x² = cos² z Step 4: Since x0, cos z> 0 We would get: x = COS z 1 sin(z) y = cos z. == sin z cos(2) Hence the parametric form is: (x, y, z) = (cos z, sin z, z), — √ <≤ z <
Continuing with this problem, shown in the picture, where the parameterization for S was determined, how do i now calculate the area/surface of S. Could you help set up the integral for this equation?
Transcribed Image Text:N
-2
-1.5
2
1.5
-1
-0.51
-0.5
-1
-1.5
-2
y
1.5
2
1.5
X
2
Transcribed Image Text:(x, y, z) = (cos z, sin z, z), − < z <
Explanation:
Step 1:
y = x tan(z)
with:
x>0, x²+ y² = 1
−√ < ± < √
Step 2:
sin(z)
y = x cos(2)
Step 3:
substitute y:
x²+x2. sin².
cos2
Then we get:
=
2 . (cos's+sin= 2)
Hence we would get:
x² = cos² z
Step 4:
Since
x0, cos z> 0
We would get:
x = COS z
1
sin(z)
y = cos z.
== sin z
cos(2)
Hence the parametric form is:
(x, y, z) = (cos z, sin z, z), — √ <≤ z <
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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