# Thread: Volume of a sphere cap

1. ## Volume of a sphere cap

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.

2. 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.
Just rotate the part of the circle $\displaystyle x^2 + y^2 = r^2$ between y = r - h and y = r around the y-axis:

$\displaystyle V = \pi \int_{y = a}^{y = b} x^2 \, dy = \pi \int_{r - h}^{r} (r^2 - y^2) \, dy = ....$

3. Originally Posted by mr fantastic
Just rotate the part of the circle $\displaystyle x^2 + y^2 = r^2$ between y = r - h and y = r around the y-axis:

$\displaystyle V = \pi \int_{y = a}^{y = b} x^2 \, dy = \pi \int_{r - h}^{r} (r^2 - y^2) \, dy = ....$
Oh that's brilliant!

Thank you, it would have taken me forever to see that.