1. ## Volumes

The base of S is a circular disk (most would call this a circle!) with radius r. Parallel cross sections perpendicular to the base are isosceles triangles with height h and unequal side of the base.
a) Set up the integral for the volume of S
b) By interpreting the integral as an area, find the volume of A.

So, I took some mushrooms and now I think that I have this visualized (more or less). The base of the triangle straddles the x-axis, so its base is 2y and its height is h. If each end of the base touches the circle, then

$y = \sqrt{r^2 - x^2)$

So, $A = \frac{1}{2} \cdot 2 yh = h\sqrt{r^2 - h^2}$

And, $V = \int_{-r}^r h\sqrt{r^2 - h^2}dx$

Am I right so far? What I don't see is how to integrate this. I can do a trig substitution for the radical, but what do I do with h? I think that this is a clue that I have done something wrong.

Can anyone help? Thanks.

2. ## Re: Volumes

Originally Posted by joatmon
The base of S is a circular disk (most would call this a circle!) with radius r. Parallel cross sections perpendicular to the base are isosceles triangles with height h and unequal side of the base.
a) Set up the integral for the volume of S
b) By interpreting the integral as an area, find the volume of A.

So, I took some mushrooms and now I think that I have this visualized (more or less). The base of the triangle straddles the x-axis, so its base is 2y and its height is h. If each end of the base touches the circle, then

$y = \sqrt{r^2 - x^2)$

So, $A = \frac{1}{2} \cdot 2 yh = h\sqrt{r^2 - h^2}$

And, $V = \int_{-r}^r h\sqrt{r^2 - h^2}dx$

Am I right so far? What I don't see is how to integrate this. I can do a trig substitution for the radical, but what do I do with h? I think that this is a clue that I have done something wrong.

Can anyone help? Thanks.
almost correct. should be ...

$V = \int_{-r}^r h\sqrt{r^2 - x^2}dx$

using symmetry and noting that $h$ is a constant ...

$V = 2h \int_0^r \sqrt{r^2 - x^2}dx$

now do your planned trig sub