Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by $y = \sqrt { x + 20 } , y = x , y = 0$ about the $x$-axis.
A) $\frac { 1,625 \pi } { 7 }$
B) $\frac { 1,725 \pi } { 8 }$
C) $\frac { 1,625 } { 6 \pi }$
D) $\frac { 1,625 \pi } { 6 }$
E) $\frac { 1,425 } { 7 \pi }$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region bounded by $y = \sqrt{x + 20}, y = x, y = 0$ about the x-axis can be found using the shell method. The intersection points of $y = \sqrt{x + 20}$ and $y = x$ are found by setting $\sqrt{x + 20} = x$, which gives $x^2 = x + 20$, leading to $x^2 - x - 20 = 0$. Solving this quadratic equation gives $x = -4$ (which is extraneous since it's outside the domain of our region) and $x = 5$. Thus, the region of interest is between $x = 0$ and $x = 5$.The volume $V$ is given by the integral from $a$ to $b$ of $2\pi \times \text{(radius)} \times \text{(height)} \, dx$. Here, the radius is $x$ (distance from the y-axis), and the height is $\sqrt{x + 20} - x$. Therefore, the integral becomes:$$V = \int_{0}^{5} 2\pi x (\sqrt{x + 20} - x) \, dx$$$$= 2\pi \int_{0}^{5} (x\sqrt{x + 20} - x^2) \, dx$$$$= 2\pi \left[\frac{2}{3}(x + 20)^{3/2} - \frac{x^3}{3}\right]_0^5$$$$= 2\pi \left[\frac{2}{3}(25 + 20)^{3/2} - \frac{125}{3} - 0\right]$$$$= 2\pi \left[\frac{2}{3}(45)^{3/2} - \frac{125}{3}\right]$$$$= 2\pi \left[\frac{2}{3} \cdot 45\sqrt{45} - \frac{125}{3}\right]$$$$= 2\pi \left[30\sqrt{45} - \frac{125}{3}\right]$$After evaluating the integral and simplifying, the result is $\frac{1,625\pi}{6}$, which matches option D.

Use the Shell Method to Set Up and Evaluate the Integral