# Evaluate the double integral using substitution:

Evaluate: $$\int_0^1\int_0^\sqrt{1-x^2}e^{-(x^2+y^2)}\,\mathrm dy\;\mathrm dx$$

I am trying to use substitution on this problem by making $x^2+y^2 = u$ and $\sqrt{1-x^2} = v$. I then tried to add $u$ and $v$ to get something like $v + u = x^2 + y^2 + \sqrt{1-x^2}$ and I'm trying to set it equal to $x$, and then $y$, but I'm not sure if I'm doing this correctly.

Any suggestions as to how to make the substitution process less tedious?

Thank you so much.

One easy way to see how to change variables is to see that you are integrating over a quarter of a circle. Then switch to polars: $x^2+y^2 = r^2 \in [0,1]$ and $\theta \in [0,\pi/2]$. Use the Jacobian $dx \, dy = r \, dr \, d\theta$. Then the integral is

$$\int_0^1 dr \, r \, e^{-r^2} \, \int_0^{\pi/2} d\theta = \frac{\pi}{4} \int_0^1 du \, e^{-u} = \frac{\pi}{4} \left ( 1-\frac{1}{e}\right)$$

Whenever you see an integral with $x^2+y^2$ and a $\sqrt{1-x^2}$ or a $\sqrt{1-y^2}$, the best substitution is to go into polar coordinates. Don't make a custom substitution unless you absolutely must. (A custom substitution would require computing the Jacobian, which isn't a trivial thing.)

Now, recall that, for polar, $r=x^2+y^2$. We can now make the substitution: $$\iint e^{-r^2}r\;\mathrm dr\;\mathrm d\theta$$

But, what are our limits of integration? To answer this, draw the region given in the question. It is the part of the unit circle in the first quadrant. Thus, our new integral is:

$$\int_0^{\pi/2}\int_0^1 e^{-r^2}r\;\mathrm dr\;\mathrm d\theta$$

I'm sure you can proceed from here.

• Would you be able to finish this? so i can see how you solved it. Thank You – jain smit Jun 29 '13 at 17:59
• @jainsmit just use substitution $t=r^2$. Result is $\frac{\pi(e-1)}{4e}$. – Ruslan Jun 29 '13 at 18:54
• @Ruslan I'm still not following you. – jain smit Jun 29 '13 at 19:16
• @jainsmit $\int_0^1 e^{-r^2}r dr=\int_0^1 \sqrt{t}\frac{1}{2\sqrt{t}}e^{-t}dt=\frac12\int_0^1 e^{-t}dt=\left. -\frac12 e^{-t}\right|_0^1=-\frac12\frac{1-e}{e}$. Since this doesn't depend on $\theta$, outer integral just multiplies this by $\pi/2$. – Ruslan Jun 29 '13 at 19:30
• so the answer would be about 0.49647 – jain smit Jun 29 '13 at 22:57