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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!

0 and 1 :)

disks

x^2+5

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 ;/

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