A Double Integral Substitution

Question

I am asked to compute the integral $$\int_\Omega xye^{x^2-y^2}dxdy$$ over the domain $\Omega = \{(x,y)\mid 1\leq x^2-y^2\leq 9, 0\leq x \leq 4, y\geq 0\}.$ After splitting the domain and a messy calculation of the resulting iterated integrals I get the answer $2e(e^8-2),$ which looks very much like what one would get after a nice substitution, but I cannot see how. I tried setting $u(x,y) = x, v(x,y) = x^2-y^2,$ and taking half of the resulting double integral but then I'm getting $2e(e^8-1).$ What am I missing?

Answer 1

I'm guessing that when you tried to invert your substitution to use it with the change of variables formula, you got

$$F(x,y)=\left(x,\sqrt{x^2-y}\right),$$ $$\det J_F =\frac{1}{2\sqrt{x^2-y}}$$

and ended up with the integral

$$ \int_1^9 \int_0^4 \frac 12 xe^y \,dx\,dy= 2 \cdot 2e(e^8-1).$$

If this is so, then you should have checked that $F$ actually sends the rectangle $[0,4] \times [1,9]$ to $\Omega$!

Let's run through the check:

On the line $x=4$, $F$ sends the point $(4, y)$ to $(4,\sqrt{16-y})$, i.e. the line $x=4$.
(Since $y \in [1, 9]$, the second component is well-defined.)

On the line $y=1$, $F$ sends the point $(x, 1)$ to $(x,\sqrt{x^2-1})$, i.e. the upper half of $x^2-1=y^2$.
We need $x \in [1, 4]$ for the second component to be well-defined.

On the line $y=9$, $F$ sends the point $(x, 9)$ to $(x,\sqrt{x^2-9})$, i.e. the upper half of $x^2-9=y^2$.
Similarly, we need $x \in [3, 4]$ for the second component to be well-defined.

Now when we come to the line $x=0$, we get $(0, \sqrt{-y})$, which is not well defined at all!

So there are several issues here.

Let's start by finding the preimage of the line $y = 0$, since that's where the problems are.

Set $F(x,y) = (x, 0)$, and we see that we're looking for the curve $x^2=y$.

After some careful consideration, we find that our new region to integrate over is the rectangle $[0,4] \times [1,9]$ intersected with the region under the parabola $y = x^2$.

That means our integral is now

$$ \int_1^3 \int_1^{x^2} \frac 12 x e^y\,dy\,dx + \int_3^4 \int_1^9 \frac 12 x e^y\,dy\,dx,$$

which evaluates to $2e(e^8-2)$, as we expect.

Answer 2

You can evaluate this integral directly. Let $\Omega=\Omega_1 \cup\Omega_2 $, where $$\Omega_1:1\leq x\leq 3,0\leq y\leq \sqrt{x^2-1},$$ $$\Omega_1:3\leq x\leq 4,\sqrt{x^2-9}\leq y\leq \sqrt{x^2-1}.$$ So $$\iint_\Omega xye^{x^2-y^2}dxdy =\iint_{\Omega_1} xye^{x^2-y^2}dxdy+\iint_{\Omega_2} xye^{x^2-y^2}dxdy.$$ Then we evaluate the two integrals respectively: $$\iint_{\Omega_1} xye^{x^2-y^2}dxdy =\int_{1}^{3}xe^{x^2}\left(\int_{0}^{\sqrt{x^2-1}}ye^{-y^2}dy\right)dx$$ $$=\frac{1}{2}\int_{1}^{3}x(e^{x^2}-1)dx=\frac{1}{4}(e^9-9e);$$ $$\iint_{\Omega_2} xye^{x^2-y^2}dxdy =\int_{3}^{4}xe^{x^2}\left(\int_{\sqrt{x^2-9}}^{\sqrt{x^2-1}}ye^{-y^2}dy\right)dx$$ $$=\frac{1}{2}\int_{3}^{4}x(e^{9}-e)dx=\frac{7}{4}(e^9-e).$$ From above, we can get : $$\iint_\Omega xye^{x^2-y^2}dxdy=2e(e^8-2).$$