Shell Method Problem: Solve x=y^(2), x=4 About the X-Axis

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Homework Statement


x=y^(2), x=4, about the x axis.

Homework Equations



2pi* integral from a to b of radius*height of function*thickness

The Attempt at a Solution


I have 2pi* integral from -2 to 2 of y*(4-y^(2)) dy but that does not make any sense. Answer comes out to be 0. The real answer is 8pi. I know how to do this with the disk method just not the shell method. Thank for the help!
 
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hi hvidales! :smile:
hvidales said:
… integral from -2 to 2 …

noooo :wink:
 
I see and thanks. However, how come for this problem: y=x^(2), y=2-x^(2), about x=1 the limits are from -1 to 1 ?
 
you mean x goes from -1 to 1?

that's because the body in that case goes from x = -1 to x = 3 (the reflection about x = 1)

in your first example the body went from y = -2 to y = 2

in each case, you're taking half (because each single cylindrical shell is in both halves) :wink:
 
Thank you!
 
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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