region $r$ is enclosed by the curve $y = 2\\sqrt{x - 3}$, the line $y = 4$, and the line $x = 3$. what is…

Explanation:

Step1: Recall the washer - method formula

The volume $V$ of a solid of revolution about the $y$-axis using the washer method is $V=\pi\int_{a}^{b}([R(y)]^{2}-[r(y)]^{2})dy$. First, we need to express $x$ in terms of $y$ for the curve $y = 2\sqrt{x - 3}$, so $x=\frac{y^{2}}{4}+3$.

Step2: Find the limits of integration

The region $R$ is bounded by $y = 2\sqrt{x - 3}$, $y = 4$ and $x = 3$. When $x = 3$, $y=0$. When $y = 4$, from $y = 2\sqrt{x - 3}$, we have $4=2\sqrt{x - 3}$, then $\sqrt{x - 3}=2$ and $x = 7$. The limits of integration for $y$ are from $y = 0$ to $y = 4$.

Step3: Determine the outer and inner radii

The outer - radius $R(y)$ is the distance from the $y$-axis to the line $x = 3$, so $R(y)=3$. The inner - radius $r(y)$ is the distance from the $y$-axis to the curve $x=\frac{y^{2}}{4}+3$, so $r(y)=\frac{y^{2}}{4}+3$.

Step4: Set up the integral

$V=\pi\int_{0}^{4}(3^{2}-(\frac{y^{2}}{4}+3)^{2})dy=\pi\int_{0}^{4}(9 - (\frac{y^{4}}{16}+\frac{3y^{2}}{2}+9))dy=\pi\int_{0}^{4}(-\frac{y^{4}}{16}-\frac{3y^{2}}{2})dy$.

Step5: Integrate term - by - term

$\int(-\frac{y^{4}}{16}-\frac{3y^{2}}{2})dy=-\frac{y^{5}}{80}- \frac{y^{3}}{2}+C$.

Step6: Evaluate the definite integral

$V=\pi\left[-\frac{y^{5}}{80}-\frac{y^{3}}{2}\right]_{0}^{4}=\pi\left(-\frac{4^{5}}{80}-\frac{4^{3}}{2}\right)=\pi\left(-\frac{1024}{80}-32\right)=\pi\left(-\frac{64}{5}-32\right)=\pi\left(\frac{- 64 - 160}{5}\right)=\frac{9031\pi}{80}$.

Answer:

$\frac{9031\pi}{80}$