Finding volume using integration

theBEAST
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Homework Statement


Given x=y^2, x=1 what is the area of the solid when the area between the two curves is rotated about x=1.

The Attempt at a Solution


I attached my solution and according to the answer key my area formula (pi-pi*y^4) is wrong. Instead they have pi*(1-y^2)^2. Can anyone explain why my area formula is wrong? I thought that it was the area of the upper boundary curve minus the area of the lower boundary curve...
 

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theBEAST said:

Homework Statement


Given x=y^2, x=1 what is the area of the solid when the area between the two curves is rotated about x=1.

The Attempt at a Solution


I attached my solution and according to the answer key my area formula (pi-pi*y^4) is wrong. Instead they have pi*(1-y^2)^2. Can anyone explain why my area formula is wrong? I thought that it was the area of the upper boundary curve minus the area of the lower boundary curve...
What method are you trying to use? Is it disks? ... or is it shells?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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