# Solids

• Jul 4th 2010, 04:01 PM
Ulysses
Solids
A solid has a circular base of radius r=4. Find in each case the volume of the solid if all intersection between the solid and a plane perpendicular to a set diameter is:

a) A square
b) A triangle rectangle isosceles which one of its sides is on the circular base.

The first I thought was in some kind of cylindroid. I know I must find an equation that I can integrate. So, the sides of the squares will be given by:

$z=2\sqrt{r^2-x^2}$

Is the integration given by: $\displaystyle\int_{-4}^{4}[2\sqrt{r^2-x^2}]^2dx=4\displaystyle\int_{-4}^{4}(r^2-x^2)dx$ for the square?

And how can I find the volume of the other solid b)?
• Jul 5th 2010, 09:06 AM
apcalculus
I am assuming this is an isosceles triangle you're referring to, and not a rectangle. The side of the isosceles triangle is the same as the side of the square that you found in part a).

The area function for the isosceles triangle, in terms of its side a, is given by

$A = base * height * \frac{1}{2} = a \frac{a\sqrt{3}}{2} * \frac{1}{2}=a^2\frac{\sqrt{3}}{4}$

I hope this helps. Good luck!