Find volume bounded by curve $x=\sqrt {25-y^2}$, and the line $x=3$, rotated about the $y$-axis My attempt: 
$$2\pi\int_0^{5} (25-y^2) dy$$ 
What's my error?

Hint:
$x = 3$ and $x = \sqrt{25-y^2}$ intersect at $y = 4$ and $-4$
$$V = \pi \int_{-4}^{4} (25-y^2 - 9)dy = \pi \int_{-4}^{4} (16-y^2)dy$$
if you evaluate the integral you get $V = \frac{256\pi}{3}$
The Shell method of solving:
Radius $= x$ and height = $2\sqrt{25-x^2}$ 
$V = 2\pi \int_{3}^{5} 2x\sqrt{25-x^2}dx = \frac{256\pi}{3}$

