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Question
Math
Posted 7 months ago

Let RR be the leftmost region in the first quadrant enclosed by the xx-axis and the curve y=sin(x)y=\sin (x).

A solid is generated by rotating RR about the line y=1y=1.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) π0π[2sin(x)sin2(x)]dx\pi \int_{0}^{\pi}\left[2 \sin (x)-\sin ^{2}(x)\right] d x
(B) π01[2sin(x)sin2(x)]dx\pi \int_{0}^{1}\left[2 \sin (x)-\sin ^{2}(x)\right] d x
(C) π01[1sin2(x)]dx\pi \int_{0}^{1}\left[1-\sin ^{2}(x)\right] d x
(D) π0π[1sin2(x)]dx\pi \int_{0}^{\pi}\left[1-\sin ^{2}(x)\right] d x
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 2
Here, Router=1R_{\text{outer}} = 1 (the distance from y=1y=1 to the x-axis) and Rinner=1sin(x)R_{\text{inner}} = 1 - \sin(x) (the distance from y=1y=1 to the curve y=sin(x)y=\sin(x))
step 3
Therefore, the volume integral becomes V=π0π[12(1sin(x))2]dxV = \pi \int_{0}^{\pi} \left[ 1^2 - (1 - \sin(x))^2 \right] dx
step 4
Simplifying the integrand: 12(1sin(x))2=1(12sin(x)+sin2(x))=2sin(x)sin2(x)1^2 - (1 - \sin(x))^2 = 1 - (1 - 2\sin(x) + \sin^2(x)) = 2\sin(x) - \sin^2(x)
step 5
Thus, the integral for the volume is V=π0π[2sin(x)sin2(x)]dxV = \pi \int_{0}^{\pi} \left[ 2\sin(x) - \sin^2(x) \right] dx
A
Key Concept
Washer Method
Explanation
The washer method is used to find the volume of a solid of revolution when the region is rotated about a line that is not the x-axis or y-axis. It involves subtracting the volume of the inner solid from the volume of the outer solid.

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