monroe17
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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tkhunny
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Find the points of intersection. This will also be your integral limits with this lovely region.
monroe17
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how do i do that?
tkhunny
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\(x^{2} = \sqrt{x}\) -- Solve!
monroe17
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0 and 1 :)
tkhunny
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Perfect.
Now, decide disks or shells. With this well-behaved region, it really doesn't much matter. Just pick one.
monroe17
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disks
tkhunny
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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?
monroe17
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x^2+5
tkhunny
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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?
monroe17
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my bad, and R right?
tkhunny
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No worries. x^2 is the inner radius. Are we calling that 'r'?
What is the outer radius, "R"?
monroe17
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isn't (-5-x^2) the outer radius? and yes
tkhunny
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No, \(x^{2}\) and \(\sqrt{x}\) are a little funny on [0,1] \(\sqrt{x} \ge x^2\;for\;x\in[0,1]\)
monroe17
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uhm.. im not sure ;/
tkhunny
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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.
monroe17
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I'm still confused toward the sqrt(x) where'd that come from?
tkhunny
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x = y^2 ==> y = sqrt(x)
We have integration "dx", so we need the representation in terms of x.
monroe17
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got it ;p lol just had to see the steps..
tkhunny
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Well, there you have it. Let's see what you get.
monroe17
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I got 109pi/30?
tkhunny
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THAT is the right answer. Good work!
Wading through it is FUN, isn't it?!
monroe17
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haha.. totally ;p
tkhunny
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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!