Region R is enclosed by the lines $y=1$, $x=4$, and the curve $y=\sqrt{x}$. Rotation around x-axis

Question

To be clear, I know how to do this problem using the washer method. I am trying to solve it using cylindrical shells. Here is what I have so far:

$2\pi \int_{1}^{2} y*y^2 dy$

$2\pi \int_{1}^{2} y^3 dy$

I get $15/2\pi$ as my final answer. the correct answer is supposed to be $9/2 * \pi$. What am I doing wrong?

Answer 1

You are calculating the wrong region, draw a picture! Note that you should be looking at $4-y^{2}$ instead and indeed $$2\pi\int_{1}^{2}y(4-y^{2}) \ dy =\frac{9\pi}{2}.$$

Answer 2

Disc Method $$ \begin{aligned} \textrm{ Volume of the hollow solid }=& \pi \int_1^4(\sqrt{x})^2 d x-3 \pi(1)^2 \\ = & \pi\left[\frac{x^2}{2}\right]_1^4-3 \pi \\ = & \frac{15 \pi}{2}- 3 \pi \\ = & \frac{9 \pi}{2} \end{aligned} $$