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Find the volume formed by rotating the region enclosed by: y=x^2, x=y^2 about the line x=-5

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Find the points of intersection. This will also be your integral limits with this lovely region.
how do i do that?
\(x^{2} = \sqrt{x}\) -- Solve!

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0 and 1 :)
Perfect. Now, decide disks or shells. With this well-behaved region, it really doesn't much matter. Just pick one.
Disks it is, then. \(\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} d(Height)\) The "Height" is the x-direction, so we have: \(\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} dx\) Now indentify that inner and outer radii. What say you?
Well, I would prefer the more literal (-5-x^2), but it will do since we are just going to square it. Is that the inner or the outer radius?
my bad, and R right?
No worries. x^2 is the inner radius. Are we calling that 'r'? What is the outer radius, "R"?
isn't (-5-x^2) the outer radius? and yes
No, \(x^{2}\) and \(\sqrt{x}\) are a little funny on [0,1] \(\sqrt{x} \ge x^2\;for\;x\in[0,1]\)
uhm.. im not sure ;/
r is \((-5 - x^{2})\) R is \((-5 - \sqrt{x})\) That's all. Did you draw a picture? You will see it at a glance.
I'm still confused toward the sqrt(x) where'd that come from?
x = y^2 ==> y = sqrt(x) We have integration "dx", so we need the representation in terms of x.
got it ;p lol just had to see the steps..
Well, there you have it. Let's see what you get.
I got 109pi/30?
THAT is the right answer. Good work! Wading through it is FUN, isn't it?!
haha.. totally ;p
Once we realized what the region was, and where the square root came from, you were all over it. This is encouraging! Move on to the next one!

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