Originally Posted by

**angel.white** Can someone help me out on this problem?

We are supposed to find the volume using integrals. Meaning as x or y changes, the cross-sectional area will change, so we find the equation for the cross sectional area, and then take the integral of it across the height.

In this one, I figured the cross-sectional area = $\displaystyle \pi x^2$

And if we set the bottom of the blue section along the x-axis, then it would be integrated from 0 to h, so

volume = $\displaystyle \pi \int_0^h x^2~dx$

volume = $\displaystyle \pi [\frac 13 x^3]_0^h$

volume = $\displaystyle \frac 13 \pi~h^3$

But somehow I'm supposed to incorporate the radius.

Back of the book says the answer is $\displaystyle \pi h^2(r-\frac 13 h)$

edit: well, part of my problem, I think is that I should be puting the cross-sectional area in terms of y. And I think I should be using the equation of a circle: x^2=r^2-y^2. That still doesn't get me the right answer, but it's much closer to theirs, as this would become $\displaystyle \pi \int_0^h (r^2-y^2)~dy = \pi ( r^2h-h^3/3)$ so getting closer :/ But this method seems like it would integrate just the top half of the circle. So somehow I need to remove that gap between the circle's center, and the bottom of h.