What is the error in finding the volume of a cap of a sphere using integration?

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Homework Statement


Find the volume of a cap of a sphere with radius r and height h.
A picture of it is given here;
http://sjc.ilrn.com/ilrn/bca/user/appletImage?dbid=1276286560



Homework Equations


The area of a circle is \pi\r^2



The Attempt at a Solution


So I flattened the sphere into a circle and solved me equation for x;
x=sqrt(r^2-y^2)
Then I rotated it around the y-axis and integrated to find;
\pi\int\stackrel{h}{r}(r²-y²)dy= \pi((1/3)h³+(2/3)r³-hr²)

But the computer program I do my homework on tells me that it wrong. Now I'm confused. What did I do wrong?
 
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As h->0 the volume should go to zero, right? Does yours? That's a clue something is wrong. Look at your limits of integration again. Are they right?
 

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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