Using the shell method, find the volume of: $y = \sqrt x$ with bounds $[0,4]$ after it has been revolved around the y axis.
After plugging everything into the shell formula I get:
$$2 \pi \int_0 ^4 x \sqrt x dx$$
After evaluating everything I get 128$\pi$/5. My textbook has the correct answer as 32$\pi$/5.
Am I making an error or is shell method just impossible in this scenario? If it is not possible, please explain why, thanks!
(note, my book did not ask to solve using shell, it asked using disk method, I am just trying shell for fun.)
Best Answer
Your answer is correct, assuming that you're looking at the right region. You can check by trying it with washers:$$V = \pi \int_0^2 (16 - y^4) dy = \pi\left(32 - \frac{32}{5} \right) = \frac{128\pi}{5}.$$ This is for the region bounded by $y = \sqrt x$, $x = 4$, and $y = 0$. If, however, the intended region was above the curve, i.e., bounded by $y = \sqrt x$, $y = 2$, and $x = 0$, then the book's answer is right, and your setup is wrong: it should be $$2\pi\int_0^4 (2 - x \sqrt x) dx.$$