Surface Integrals — Question 9

Question 9 – Solution

Strategy. On this paraboloid, the denominator in the density is exactly the graph-area factor, leaving a Gaussian radial integral.

Step 1: Truncate Let $S_R$ be the part above $0\le r\le R$. For $z=r^2$, $$dS=\sqrt{1+4r^2}\,r\,dr\,d\theta.$$ Therefore the density times the area element simplifies to $$\frac{e^{-r^2}}{\sqrt{1+4r^2}} \sqrt{1+4r^2}\,r\,dr\,d\theta =e^{-r^2}r\,dr\,d\theta.$$

Step 2: Evaluate the truncation $$\begin{align*} \iint_{S_R}\frac{e^{-z}}{\sqrt{1+4z}}\,dS &=2\pi\int_0^R e^{-r^2}r\,dr\\ &=2\pi\left[-\frac 12e^{-r^2}\right]_0^R\\ &=\pi(1-e^{-R^2}). \end{align*}$$

Step 3: Take the limit The integrand is nonnegative and the truncated values increase to a finite limit: $$\boxed{\iint_S\frac{e^{-z}}{\sqrt{1+4z}}\,dS =\lim_{R\to\infty}\pi(1-e^{-R^2})=\pi}.$$

Verification The omitted tail equals $\pi e^{-R^2}$, which tends to zero and quantifies the convergence.