Question

Find the volume of the solid whose base is the circle x2+y2=25 and the cross sections perpendicular to the x-axis are triangles whose height and base are equal. Find the area of the vertical cross section A at the level x=4.

Answer 1

Triangles with height $h$ and base $b$, with $b=h$ have area $\dfrac{b^2}2$.

Such cross sections with the base of the triangle in the disk $x^2+y^2\le25$ (a disk with radius 5) have base with length

$b(x)=\sqrt{25-x^2}-\left(-\sqrt{25-x^2}\right)=2\sqrt{25-x^2}$

i.e. the vertical (in the $x,y$ plane) distance between the top and bottom curves describing the circle $x^2+y^2=25$.

So when $x=4$, the cross section at that point has base

$2\sqrt{25-16}=6$

so that the area of the cross section would be 6^2/2 = 18.

In case it's relevant, the entire solid would have volume given by the integral

$\displaystyle\int_{-5}^5\frac{b(x)^2}2\,\mathrm dx=4\int_0^5(25-x^2)\,\mathrm dx=\frac{1000}3$