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ruslelena [56]
3 years ago
6

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 triangle

s whose height and base are equal.
Mathematics
1 answer:
netineya [11]3 years ago
5 0
Because the height and base of each cross section are equal, the area for any given cross section is \dfrac12bh=\dfrac12b(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\sqrt{64-x^2}

so that

b(x)=\sqrt{64-x^2}-(-\sqrt{64-x^2})=2\sqrt{64-x^2}

and so the area of each cross section is

\dfrac12(2\sqrt{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=\dfrac{4096}3
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