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Find the volume of the solid whose base is the circle x^2+y^2=64 and the cross sections perpendicular to the x-axis are triangles whose height and base are equal.

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Because the height and base of each cross section are equal, the area for any given cross section is
\frac12bh=\frac12b(x)^2 where the base of each section occurring along the line
x=x_0 is the vertical distance between the upper and lower halves of the circle
x^2+y^2=64.

We can write


y=\pm√(64-x^2)

so that


b(x)=√(64-x^2)-(-√(64-x^2))=2√(64-x^2)

and so the area of each cross section is


\frac12(2√(64-x^2))^2=2(64-x^2)=128-2x^2

Over the circular base of the solid, we have
-8\le x\le8, so the volume of the solid is given by the integral


\displaystyle\int_(x=-8)^(x=8)(128-2x^2)\,\mathrm dx=\frac{4096}3
User Bartosz Firyn
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