Question

Find the volume of the solid under z=2x-y and above D which is bounded by the circle with center at the origin of radius R in the xy plane
I know this is Calc 3 not diffyq but I'm stumped. I keep getting 0

Answer 1

What was your process leading up to your answer? I have not done much with this topic, but I will see if I can be of any use.

Answer 2

here is my work that i've got on it. I started by putting the cordinates into cylindrical form and from there found what my limits of integration for D would be. I think the limits for z are right however i might be wrong. After that its just simple integration, but no matter what I keep getting zero which to me just does not seem to fit

Answer 3

I am not finding the issue here at the moment. My apologies, I was not able to be of use with this! I would next be able to take a shot at the problem tomorrow, which is generally where help becomes unnecessary. ..

I recommend bumping this question or tagging some others who are online that may be more capable in this subject. :)

Answer 4

The problem is that z=2x-y goes below the xy-plane so the signed volume would be 0 (positive and negative would cancel). However, the problem states that the boundary is 0<=z<=2x-y which is the space above the xy-plane, so you must account for this by figuring out which regions fit this description.

Answer 5

So we have\[z=2x-y \ge0\] and converting this into polar we get\[z=2rcos \theta-rsin \theta \ge0\]solving for theta and remembering the domain of tangent:\[2 \ge \tan \theta\]\[-\frac{ \pi }{ 2 }\le \theta \le \tan ^{-1}(2)\] Using these limits of integration for theta should give you a non-zero volume.

Answer 6

Remember to use this triangle when evaluating the sine and cosine for your final answer.

Answer 7

There should be a factor of 2 multiplied into the integrand as well to account for the other half of the circle as well.