Originally Posted by

**arze** I have two questions here, the first i need someone to show me my mistake somewhere, the second I have no idea how to do.

1. The area defined by the inequalities

$\displaystyle 0\geq y\geq x^2$, $\displaystyle 0\geq x\geq 2$

is rotated about the y-axis. Find the volume generated.

I take the raduis of the figure generated from the y-axis outward. so then its value would be $\displaystyle x_{further}-x_{nearer}=2-\sqrt{y}$

This rotates the region under the curve, you should be rotating the region above it.

$\displaystyle V=\int^4_0 \pi x^2 dy$

$\displaystyle V=\pi \int^4_0(2-\sqrt{y})^2 dy$

$\displaystyle V=\pi \int^4_0(4-4\sqrt{y}+y) dy$

$\displaystyle V=\pi [4y-\frac{8}{3}y^{\frac{3}{2}}+\frac{1}{2}y^2]^4_0$

$\displaystyle V=\frac{8\pi}{3}$

answer is $\displaystyle 8\pi$

2. Find the volume generated when the area in the first quadrant enclosed by $\displaystyle y=|x^2-1|$, and the line y=1 is rotated about the line y=1.

I only know that the lines intersect at x=0 and $\displaystyle \sqrt{2}$. Also the radii of the figure would be $\displaystyle 1-|x^2-1|$.

Thanks!