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I have the area [tex]D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1[/tex]

That is rotated about the x axis, and i need to calculate the area

[tex]\pi \int_0^1 3^2-y^2 = \pi \int_0^1 9-xe^{2x^2}[/tex]

[tex]\frac{-9\pi}{4}\cdot (e^{2x^2}-1)\bigg|_0^1[/tex]

But this is all wrong, why?

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# Homework Help: Solids of revolution

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