0

Evaluate the Surface Integral $xyz$ $dS$ where $S$ is the surface defined by $2y=\sqrt{9-x}$, $x>0$, and between $z=0$ and $z=3$.

Don't even know where to start with this question.

JKnecht
  • 6,543
Mike
  • 39
  • An idea: begin by parametrizing your surface, say $$r(u,v)=\left(u,,\frac{\sqrt{9-u}}2,,v\right);,;;0\le u\le 9;,;;0\le v\le 3$$ and now do the vectorial product $$r'_u\times r'_v$$ and take its norm. Finally, according to your parameters intervals, your integral is $$\int\int_A x(u,v)y(u,v)z(u,v)\cdot ||r'_u\times r'_v|| dA$$ with $;A;$ the zone on the $;uv-$plane determined by the parameters and $;x,y,z;$ are given according to the parametrization $;r(u,v);$ – DonAntonio Feb 08 '16 at 00:06
  • When it comes to Mathjax, generally two dollar signs should be next to each other only to put math in display mode (i.e., on its own line). So rather than $xyz$ $dS, $xyz\ dS$ should be accomplished with some kind of whitespace command, like xyz\ dS, xyz\,dS, or $xyz~dS$, things like that @JKnecht – pjs36 Feb 08 '16 at 02:25

0 Answers0