gsingh2011
Reputation
327
Top tag
Next privilege 500 Rep.
Access review queues
Badges
2 16
Newest
Impact
~15k people reached

• 0 posts edited
• 1 helpful flag
• 25 votes cast

# 59 Actions

 Feb6 awarded Citizen Patrol Jul2 awarded Curious Apr11 awarded Popular Question Mar3 awarded Popular Question Jan28 awarded Notable Question Dec18 accepted Invariant determinant in change of basis Dec18 asked Invariant determinant in change of basis Nov18 comment Why can you turn clothing right-side-out? @rschwieb There should be a badge for that... Nov18 awarded Popular Question Aug29 awarded Popular Question May13 awarded Caucus Feb21 awarded Popular Question Dec3 awarded Teacher Aug31 awarded Yearling May1 awarded Nice Question Apr12 accepted Surface Integral of $xy$ over the surface $y^2+z^2=36$ Apr12 comment Evaluating the integral $\int x\,dV$ where $V$ is the region bounded the surface $x^2+y^2+z^2=1$ and the planes $x = 0$, $y = x$, $z=0$ @ZevChonoles I think you mean phi goes from 0 to pi/2. Theta should be between 0 and pi/4 correct? Apr10 comment Surface Integral of $xy$ over the surface $y^2+z^2=36$ So first of all, your upper bound for x is always negative, so it should really be a lower bound. But x should never go below zero because it's in the first quadrant... Apr10 awarded Critic Apr10 comment Surface Integral of $xy$ over the surface $y^2+z^2=36$ Oh, I think I see. The maximum value of y is when x=0. Then $0+y^2=25$, so $6\cos(\theta)=5$. Thanks for the answer, I'll get back to you after I complete the problem.