Surface Integrals — Question 2

Question 2 – Solution

Strategy. Use the graph formula first, then exploit radial symmetry so the surface integral becomes one one-variable integral.

Step 1: Surface element With $g(x,y)=x^2+y^2$, $$dS=\sqrt{1+4x^2+4y^2}\,dA.$$ In polar coordinates, $z=r^2$ and $dA=r\,dr\,d\theta$, so $$\boxed{dS=\sqrt{1+4r^2}\,r\,dr\,d\theta}.$$

Step 2: Integrate $$\begin{align*} \iint_S z\,dS &=2\pi\int_0^1 r^3\sqrt{1+4r^2}\,dr. \end{align*}$$ Let $t=1+4r^2$. Then $r\,dr=dt/8$ and $r^2=(t-1)/4$, giving $$\begin{align*} \iint_S z\,dS &=\frac{\pi}{16}\int_1^5(t^{3/2}-t^{1/2})\,dt\\ &=\frac{\pi}{16}\left[\frac 25t^{5/2}-\frac 23t^{3/2}\right]_1^5\\ &=\boxed{\frac{\pi(25\sqrt 5+1)}{60}}. \end{align*}$$

Verification The integrand is nonnegative. Also $0\le z\le 1$, so the answer must not exceed the surface area; numerically it is about $2.98$, consistent with that bound.