3. Use double integral to find the volume of the cylinder-like object that exists above the first quadrant on the xy plane and below the plane z = 10. The cross section of this "cylinder" on the xy plane is given by the polar equation r = 50 sin (20) (shown in figure, ignore the scale/numbers). Use the formula Volume = zdA, R where R is the domain over which we are interested to find the volume under the surface z. Sketch the region R before doing the integration. (In polar coordinates, dA = rdrd0) 0.75 0.5 0.2 -0.75 -0.5 0.25 0.5 0.75 -0.2 0.5 -0.75 Note that this "cylinder" is not the "usual circular" cylinder that we are familiar with
3. Use double integral to find the volume of the cylinder-like object that exists above the first quadrant on the xy plane and below the plane z = 10. The cross section of this "cylinder" on the xy plane is given by the polar equation r = 50 sin (20) (shown in figure, ignore the scale/numbers). Use the formula Volume = zdA, R where R is the domain over which we are interested to find the volume under the surface z. Sketch the region R before doing the integration. (In polar coordinates, dA = rdrd0) 0.75 0.5 0.2 -0.75 -0.5 0.25 0.5 0.75 -0.2 0.5 -0.75 Note that this "cylinder" is not the "usual circular" cylinder that we are familiar with
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:3. Use double integral to find the volume of the cylinder-like object that exists
above the first quadrant on the xy plane and below the plane z =
cross section of this "cylinder" on the xy plane is given by the polar equation
r = 50 sin (20) (shown in figure, ignore the scale/numbers). Use the formula
10. The
/ zdA,
Volume =
R
where R is the domain over which we are interested to find the volume under
the surface z. Sketch the region R before doing the integration. (In
polar coordinates, dA = rdrd0)
Y
0.75
0.5
0.2
-0.75
-0.5
0.25
0.25
0.5
0.75
-0.2
0.5
-0.75
Note that this "cylinder" is not the "usual circular" cylinder that
we are familiar with
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