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.

Triangles with height [tex]h[/tex] and base [tex]b[/tex], with [tex]b=h[/tex] have area [tex]\dfrac{b^2}2[/tex].Such cross sections with the base of the triangle in the disk [tex]x^2+y^2\le25[/tex] (a disk with radius 5) have base with length[tex]b(x)=\sqrt{25-x^2}-\left(-\sqrt{25-x^2}\right)=2\sqrt{25-x^2}[/tex]i.e. the vertical (in the [tex]x,y[/tex] plane) distance between the top and bottom curves describing the circle [tex]x^2+y^2=25[/tex].So when [tex]x=4[/tex], the cross section at that point has base[tex]2\sqrt{25-16}=6[/tex]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[tex]\displaystyle\int_{-5}^5\frac{b(x)^2}2\,\mathrm dx=4\int_0^5(25-x^2)\,\mathrm dx=\frac{1000}3[/tex]