Let $R$ be the region enclosed by the $x$-axis, the $y$-axis, the line $x=\pi$ and the curve $y=\sin (x)$.

A solid is generated by rotating $R$ about the $x$-axis.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) $\int_{0}^{\pi} \sin ^{2}(x) d x$
(B) $\pi \int_{0}^{\pi} \sin ^{2}(x) d x$
(C) $\pi \int_{0}^{1} \sin ^{2}(x) d x$
(D) $\int_{0}^{1} \sin ^{2}(x) d x$

Question

Let $R$ be the region enclosed by the $x$-axis, the $y$-axis, the line $x=\pi$ and the curve $y=\sin (x)$.

A solid is generated by rotating $R$ about the $x$-axis.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) $\int_{0}^{\pi} \sin ^{2}(x) d x$
(B) $\pi \int_{0}^{\pi} \sin ^{2}(x) d x$
(C) $\pi \int_{0}^{1} \sin ^{2}(x) d x$
(D) $\int_{0}^{1} \sin ^{2}(x) d x$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the volume of the solid generated by rotating the region $R$ about the $x$-axis, we use the disk method. The formula for the volume is given by $V = \pi \int_{a}^{b} [f(x)]^2 dx$. ‖ ‖ step 2 ⋮ Here, $f(x) = \sin(x)$, $a = 0$, and $b = \pi$. So, the volume is $V = \pi \int_{0}^{\pi} \sin^2(x) dx$. ‖ ‖ step 3 ⋮ Therefore, the correct integral is $V = \pi \int_{0}^{\pi} \sin^2(x) dx$. ‖ ***B*** ∻Key Concept∻ ⚹Disk Method⚹ ∻Explanation∻ ⚹The disk method is used to find the volume of a solid of revolution by integrating the square of the function being rotated around the axis of rotation.⚹◊What is the formula for the volume of a solid generated by rotating a region enclosed by a curve around the x-axis?⍭ Generate me a similar question◊

Sia