# polar coordinates for integral bounds with parallelogram as region

I'm having trouble with this problem: Evaluate the following double integral

$$\iint_D (x^2+y^2) \,\text d x\,\text dy$$

where $D$ is the region comprised of a parallelogram with the corners $(0,0)$, $(-1,1)$, $(4,2)$ and $(3,3)$.

If I look at the function, it seems easier to solve with polar coordinates, but if I look at the region, it seems easier to do in the cartesian plane. The cartesian plane makes the calculation very complex with many steps and I seem to always make a mistake when trying it.

I've figured that the region is comprised of the lines $A: x=2y$, $B: x= 2y-3$, $C: x=-y$ and $D: x=-y+6$.

I would transform the function to polar form as follows:

$$\iint_Dr^2 \, \text dr \, \text d\theta$$

but how do I switch the integral bounds to polar form? I tried to simply substitute $x = r\cos\theta$ and $y=r\sin\theta$ for the lines $A$, $B$, $C$ and $D$; but I don't know if I should try solving for $\theta$ or for $r$, and I don't feel very confident with the trigonometry when I test this. Maybe I can use the Jacobian matrix to make it easier for me, but I haven't used the Jacobian matrix in practice before, so I feel a bit lost here.

• is this integral easier with polar coordinates?
• how do you change to polar coordinates for these bounds?
• ...change the parallelogram into a rectangle by a linear transformation, no polar coordinates. Commented Aug 8, 2018 at 12:42
• You are computing the moment of inertia of a parallelogram with respect to a corner. You may compute the moment of inertia with respect to the centroid in a simple way (it is enough to recall the moment of inertia of a rectangle with respect to its centroid), then invoke the parallel axis theorem. Commented Aug 8, 2018 at 17:40

HINT

We can set

• $$u=x+y \implies 0\le u\le 6$$

• $$v=2y-x\implies 0\le v\le 3$$

then

$$\iint_D (x^2+y^2) \,dx\, dy=\iint_R \frac13 \left[\left(\frac{2u-v}3\right)^2+\left(\frac{u+v}3\right)^2\right] \,du\, dv$$

Edit

The Jacobian for the change of variables is

$$\left|\begin{matrix} \frac{\partial x(u,v)}{\partial u} &\frac{\partial x(u,v)}{\partial v} \\ \frac{\partial y(u,v)}{\partial v} &\frac{\partial y(u,v)}{\partial v} \end{matrix}\right|=\left|\begin{matrix} \frac23 &\frac13 \\ -\frac13 &\frac13 \end{matrix}\right|=\frac 13$$

• Where does the extra factor of $\frac{1}{3}$ on the right hand side come from? Just beginning to learn this stuff. Commented Dec 8, 2023 at 8:01
• @MaplePanda It comes out from the Jacobian determinant, see the edit for details.
– user
Commented Dec 8, 2023 at 14:46