let r be the leftmost region in the first quadrant enclosed by the x - axis and the curve y = sin(x). a…

let r be the leftmost region in the first quadrant enclosed by the x - axis and the curve y = sin(x). a solid is generated by rotating r about the line y = 1. which one of the definite integrals gives the volume of the solid? choose 1 answer: a π∫₀^π 2sin(x) - sin²(x)dx b π∫₀^π 1 - sin²(x)dx c π∫₀¹ 1 - sin²(x)dx d π∫₀¹ 2sin(x) - sin²(x)dx

let r be the leftmost region in the first quadrant enclosed by the x - axis and the curve y = sin(x). a solid is generated by rotating r about the line y = 1. which one of the definite integrals gives the volume of the solid? choose 1 answer: a π∫₀^π 2sin(x) - sin²(x)dx b π∫₀^π 1 - sin²(x)dx c π∫₀¹ 1 - sin²(x)dx d π∫₀¹ 2sin(x) - sin²(x)dx

Answer

Explanation:

Step1: Recall the disk - washer method

The volume $V$ of a solid of revolution about the horizontal line $y = k$ using the disk - washer method is given by $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer radius and $r(x)$ is the inner radius. Here, we are rotating the region $R$ about $y = 1$. The outer radius $R(x)=1$ and the inner radius $r(x)=1 - \sin(x)$.

Step2: Calculate the volume formula

The volume $V$ of the solid of revolution is $V=\pi\int_{0}^{\pi}([1]^{2}-[1 - \sin(x)]^{2})dx$. Expand $[1 - \sin(x)]^{2}$ using the formula $(a - b)^{2}=a^{2}-2ab + b^{2}$, where $a = 1$ and $b=\sin(x)$. So, $[1 - \sin(x)]^{2}=1-2\sin(x)+\sin^{2}(x)$. Then $V=\pi\int_{0}^{\pi}(1-(1 - 2\sin(x)+\sin^{2}(x)))dx$.

Step3: Simplify the integrand

Simplify $1-(1 - 2\sin(x)+\sin^{2}(x))$: [ \begin{align*} 1-(1 - 2\sin(x)+\sin^{2}(x))&=1 - 1+2\sin(x)-\sin^{2}(x)\ &=2\sin(x)-\sin^{2}(x) \end{align*} ] So the volume is $V = \pi\int_{0}^{\pi}[2\sin(x)-\sin^{2}(x)]dx$.

Answer:

A. $\pi\int_{0}^{\pi}[2\sin(x)-\sin^{2}(x)]dx$