find the volume of the solid generated by revolving the region bounded by the graph y = √x, y = 0, and x = 4…

Explanation:

Step1: Recall the washer - method formula for volume about a vertical line

The formula for the volume $V$ of the solid of revolution about the vertical line $x = a$ using the washer method is $V=\pi\int_{c}^{d}([R(y)]^{2}-[r(y)]^{2})dy$. First, we need to express $x$ in terms of $y$. Given $y = \sqrt{x}$, then $x=y^{2}$. The limits of integration for $y$ are from $y = 0$ to $y = 2$ (since when $x = 4$, $y=\sqrt{4}=2$). The outer - radius $R(y)$ is the distance from the axis of rotation $x = 5$ to the left - most curve of the region, and the inner - radius $r(y)$ is the distance from the axis of rotation $x = 5$ to the right - most curve of the region.

Step2: Determine the outer and inner radii

The outer - radius $R(y)=5 - 0=5$ (distance from $x = 5$ to the $y$ - axis), and the inner - radius $r(y)=5 - y^{2}$ (distance from $x = 5$ to the curve $x = y^{2}$).

Step3: Set up the integral

The volume $V=\pi\int_{0}^{2}([5]^{2}-[5 - y^{2}]^{2})dy=\pi\int_{0}^{2}(25-(25 - 10y^{2}+y^{4}))dy$.

Step4: Simplify the integrand

$V=\pi\int_{0}^{2}(25 - 25+10y^{2}-y^{4})dy=\pi\int_{0}^{2}(10y^{2}-y^{4})dy$.

Step5: Integrate term - by - term

Using the power rule for integration $\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int(10y^{2}-y^{4})dy=10\times\frac{y^{3}}{3}-\frac{y^{5}}{5}+C$.

Step6: Evaluate the definite integral

$V=\pi\left[\frac{10y^{3}}{3}-\frac{y^{5}}{5}\right]_{0}^{2}=\pi\left(\frac{10\times2^{3}}{3}-\frac{2^{5}}{5}\right)=\pi\left(\frac{80}{3}-\frac{32}{5}\right)$.

Step7: Find a common denominator and simplify

$V=\pi\left(\frac{400 - 96}{15}\right)=\pi\times\frac{304}{15}\approx\frac{3.14159\times304}{15}\approx63.677$.

Answer:

$63.677$