suppose that $r$ is the finite region bounded by $y = x$, $y = x + 5$, $x = 0$, and $x = 4$. find the exact…

Explanation:

Step1: Use the disk - washer method formula

The volume $V$ of the solid of revolution about the $x$-axis using the disk - washer method is $V=\pi\int_{a}^{b}([f(x)]^{2}-[g(x)]^{2})dx$, where $f(x)$ is the outer - radius function and $g(x)$ is the inner - radius function, and $[a,b]$ is the interval of integration. Here, for $0\leq x\leq4$, the outer - radius function is $y = x + 5$ and the inner - radius function is $y=x$. So, $V=\pi\int_{0}^{4}((x + 5)^{2}-x^{2})dx$.

Step2: Expand the integrand

Expand $(x + 5)^{2}-x^{2}$: [ \begin{align*} (x + 5)^{2}-x^{2}&=(x^{2}+10x + 25)-x^{2}\ &=10x + 25 \end{align*} ] So, $V=\pi\int_{0}^{4}(10x + 25)dx$.

Step3: Integrate term - by - term

Integrate $10x+25$ with respect to $x$: [ \begin{align*} \int(10x + 25)dx&=10\int xdx+25\int dx\ &=10\times\frac{x^{2}}{2}+25x+C\ &=5x^{2}+25x+C \end{align*} ]

Step4: Evaluate the definite integral

[ \begin{align*} V&=\pi\left[5x^{2}+25x\right]_{0}^{4}\ &=\pi((5\times4^{2}+25\times4)-(5\times0^{2}+25\times0))\ &=\pi(5\times16 + 100-0)\ &=\pi(80 + 100)\ &=180\pi \end{align*} ]

Answer:

$180\pi$