Question: Let $S$ be the closed surface forming the boundary of the region $V$ bounded by $x^2+y^2=3$, $z=0,\ z=6$. A vector field $\vec{F}$ is defined over $V$ with $\nabla.\vec{F}=2y+z+1$. What is the value of $\displaystyle\frac{1}{\pi} \iint_S{\vec{F}. \hat{n}dS} $ ?

Attempt: We calculate $\displaystyle \frac{1}{\pi}\int_{-\sqrt{3}}^{+\sqrt{3}}\int_{-\sqrt{3-x^2}}^{+ \sqrt{3-x^2}}\int_0^6 {\nabla.\vec{F}dzdydx}$ $=$$\displaystyle \frac{1}{\pi}\int_{-\sqrt{3}}^{+\sqrt{3}}\int_{-\sqrt{3-x^2}}^{+ \sqrt{3-x^2}}12(2+y)dydx$. Converting to polar:

$\displaystyle \frac{1}{\pi}\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}12(2+rsin(\theta))rdrd\theta=72$.

Did I make any mistake? The answer provided is $24.61$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.