So my task is to calculate a volume of a region between a plane and a paraboloid over the circle centered at $S(1,2)$ with a radius of $r = 3$.

I've found a good expression for the region, but i'm having a hard time finding the boundaries for the double integral.

I can express the circle as $(x-1)^2 + (y-2)^2 = 9$, but i'm not sure where to go from here. I'm looking for the boundaries to $r$ and $\theta$. Any hints would be great!

Thanks in advance

Solved by using substitution as Saibal pointed out.

  • 2
    $\begingroup$ Shift the origin, i.e. define new variables $u = x - 1$, $v = y - 2$, recast the whole problem in terms of $u$ and $v$ $\endgroup$
    – Saibal
    Commented Feb 29, 2016 at 6:53
  • $\begingroup$ Yes thanks alot, that did indeed do the trick! $\endgroup$
    – Thomas
    Commented Feb 29, 2016 at 7:28

1 Answer 1


Alternatively, directly with polar coordinates: $$ D=\{(r,\theta)\;|\; 0\le \theta \le 2\pi, 0\le r \le cos\theta+2sin\theta+\sqrt{-3cos^2\theta+4cos\theta sin\theta+8}\;\} $$ Although it looks like the integral we be hard to compute. But if you just need to express the problem and not solve it, this should be correct.

The bounds for $\theta$ are straightforward, and the upper bound for $r$ is obtained by replacing $x=r\cos\theta$ and $y=r\sin\theta$ in the equation of the circle. Solving for $r$ in the equation gives you the upper bound.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .