contestada

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.

Respuesta :

LammettHash
Because the height and base of each cross section are equal, the area for any given cross section is [tex]\dfrac12bh=\dfrac12b(x)^2[/tex] where the base of each section occurring along the line [tex]x=x_0[/tex] is the vertical distance between the upper and lower halves of the circle [tex]x^2+y^2=64[/tex].

We can write

[tex]y=\pm\sqrt{64-x^2}[/tex]

so that

[tex]b(x)=\sqrt{64-x^2}-(-\sqrt{64-x^2})=2\sqrt{64-x^2}[/tex]

and so the area of each cross section is

[tex]\dfrac12(2\sqrt{64-x^2})^2=2(64-x^2)=128-2x^2[/tex]

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

[tex]\displaystyle\int_{x=-8}^{x=8}(128-2x^2)\,\mathrm dx=\dfrac{4096}3[/tex]

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