I have to find the volume of the solid whose area is bound by the equations [tex]y=-x+3 \ and \ y=x^2-3x[/tex] as it revolves around the x and y axis. I approached it by finding the center of mass and then using Pappus's theorem:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]M_{x}=\int^{3}_{-1} \frac{1}{2}((-x+3)^2-(x^2-3x)^2)\,dx = \frac{64}{15}[/tex]

[tex]M_{y}=\int^{3}_{-1} x(-x+3-x^2+3x)\,dx = \frac{32}{3}[/tex]

[tex]A=\int^{3}_{-1} -x+3-x^2+3x \, \ dx = \frac{32}{3}[/tex]

So the center of mass coordinates are:

[tex]\overline{x} = \frac{M_{y}}{A} = \frac{\frac{32}{3}}{\frac{32}{3}} = 1[/tex]

[tex]\overline{y} = \frac{M_{x}}{A} = \frac{\frac{64}{15}}{\frac{32}{3}} = \frac{2}{5}[/tex]

Revolving around x:

[tex]d=(2\pi)(\frac{2}{5}) = \frac{4\pi}{5}[/tex]

[tex]V=dA=(\frac{4\pi}{5})(\frac{32}{3}) = \frac{128\pi}{15} \approx 26.8083[/tex]

Revolving around y:

[tex]d=(2\pi)(1) = 2\pi[/tex]

[tex]V=dA=(2\pi)(\frac{32}{3}) = \frac{64\pi}{3} \approx 67.0206[/tex]

However the answers in the book are 63.9848 and 72.9478 respectively. Can anyone tell me where I went wrong or if I am even approaching the problem correctly? Thank you.

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# Homework Help: Volume by revolution

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