Surface area of a paraboloid

[micke]

Hi! Could someone please take a look at my work on this problem? I get a rather complicated integral so I suspect something's wrong.

The problem:
Calculate the area of the part of the paraboloid $\displaystyle 4z = x^2 + y^2$ that lies between the cylinder $\displaystyle z = y^2$ and the plane $\displaystyle z = 3$.

My work:
A quarter of the area lies in the domain 0 <= x <= a, 0 <= y <= b(x)

a is the x-value where the paraboloid intersects the plane for y = 0: $\displaystyle x^2/4 = 3 \Rightarrow x = \sqrt {12} = a$

b(x) is the y-value where the paraboloid intersects the cylinder for every x: $\displaystyle \frac{{x^2 + y^2 }}{4} = y^2 \Rightarrow y = x/\sqrt{3} = b(x)$

The area element of the paraboloid is $\displaystyle \frac{\sqrt{x^2 + y^2 + 4}}{2}$

So I have this integral:

$\displaystyle 2\int\limits_0^{\sqrt {12} } {dx} \int\limits_0^{x/\sqrt 3 } {\sqrt {x^2 + y^2 + 4} dy}$

I can't find the antiderivative of the integrand. Switching to polar coordinates would make it easy, but I don't think the domain allows that switch.

 

[royhaas]

Let $\displaystyle y = \sqrt{x^2+4}\ tan \theta$.

 

[micke]

I don't know what I was thinking. Of course I can use polar coordinates.
$\displaystyle r=\sqrt{12}$ and since $\displaystyle y=\sqrt{3}$ at the intersection of all three surfaces I get $\displaystyle \theta=arcsin(\sqrt{3}/r) = \pi/6$