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.
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.
