I agree with the equation of the line. If you're going to use the shell method, then:
[MATH]dV=2\pi y(h-x)\,dy=2\pi y\left(h-\frac{h}{r}y\right)\,dy=2\pi \frac{h}{r} y(r-y)\,dy[/MATH]
Hence:
[MATH]V=2\pi \frac{h}{r}\int_0^r y(r-y)\,dy=2\pi \frac{h}{r}\int_0^r ry-y^2\,dy=2\pi \frac{h}{r}\left[r\frac{y^2}{2}-\frac{y^3}{3}\right]_0^r=2\pi \frac{h}{r}\cdot\frac{r^3}{6}=\frac{\pi}{3}hr^2[/MATH]
I'd actually use the disk method though, where:
[MATH]dV=\pi y^2\,dx=\pi\left(\frac{r}{h}x\right)^2\,dx=\pi\frac{r^2}{h^2}x^2\,dx[/MATH]
Hence:
[MATH]V=\pi\frac{r^2}{h^2}\int_0^h x^2\,dx=\frac{\pi r^2}{3h^2}\left[x^3\right]_0^h=\frac{\pi r^2}{3h^2}\cdot h^3=\frac{\pi}{3}hr^2[/MATH]