Solving Surface Equation Homework

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Homework Statement



attachment.php?attachmentid=37177&stc=1&d=1310620954.jpg


Homework Equations





The Attempt at a Solution



It seems very obvious that after rotation, it becomes a circle, so how to write it out mathematically?

That I don't know the general equation of a surface.
Is y^2+x^2=f^2(x) the answer?


attachment.php?attachmentid=37178&stc=1&d=1310620954.jpg
 

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Yes, this is the general equation for a circle

f(x)²=x²+y²=r²

As seen in the figure, f(x)=radius

And from polar coordinates you can get the equation:

x=r*cos(\Theta)
y=r*sin(\Theta)
x²+y²=r²(cos²(\Theta)+sin²(\Theta))=r²
 

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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