# Double Integrals in Polar Coordinates; Values of Theta

Change to an equivalent polar integral and evaluate

$$\int_0^1\Biggr(\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \frac{6}{(1+x^2+y^2)^2} dy \Biggr)dx$$

I am working on this double integral and have run into a little snag. Obviously, our best bet here is to convert to polar coordinates. Our restrictions here give us a graph that looks like a full circle with a radius of $$1$$; the $$x$$ restriction slices it in half through the $$y$$-axis. So in polar form, our new bounds would be from $$0$$ to $$1$$ for our $$r$$-values and $$\pi/2$$ to $$-\pi/2$$ for our $$\theta$$ bound. The system solved its self quite well using this method. But I have noticed that others on various websites are restricting their $$\theta$$ values from $$0$$ to $$2\pi$$.

So my question here is what should the bounds be for the $$\theta$$ variable?

• You're right, in this case $\theta \in [-\pi/2,\pi/2]$. – user275377 Apr 11 '18 at 0:40

As Alex mentioned, what you have worked out is correct, $\theta$ should be in the interval $[-\pi/2, \pi/2]$.

Let $D=\{(x,y)\in\Bbb R^2|0 \leq x \leq 1, -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}\}$

By Fubini's Theorem we have that:

$$\int_0^1\Biggr(\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \frac{6}{(1+x^2+y^2)^2} dy \Biggr)dx=\iint_D \frac{6}{(1+x^2+y^2)^2} dA$$

Now by the Change of Variables Theorem we change to polar coordinates

$$\iint_D \frac{6}{(1+x^2+y^2)^2} dA=\iint_{[\pi/2,\pi/2]\times[0,1]} \frac{6}{(1+r^2)^2} rdrd\theta$$

Now we apply Fubini's Theorem again and then simply evaluate the integral

$$\int_{\theta=-\pi/2}^{\pi/2}\Biggr(\int_{r=0}^1 \frac{6}{(1+r^2)^2} rdr\Biggr)d\theta=\int_{\theta=-\pi/2}^{\pi/2} \frac{3}{2} d\theta=\frac{3\pi}{2}$$

Note that because of the symmetry of the function $f(x,y) = \frac{6}{(1+x^2+y^2)^2}$ we could have used the limits $0$ and $2\pi$ for $\theta$, however we should remember to multiply the answer by $\frac{1}{2}$ because otherwise we would have calculated double the volume required.