# Calculating double integral by converting to polar coordinates

Question: Evaluate the integral

$$\int_0^1\int_x^1 \arctan\left(\frac{y}{x}\right) \ dy\,dx$$

My attempt:

So I've converted the integral into polar coordinates, getting the integral

$$\int_0^\frac{\pi}{4}\int_0^\frac{1}{\cos(\theta)} \theta r\,dr\,d\theta \ =\int_0^\frac{\pi}{4} \frac{\theta }{2\cos^2(\theta)} \, d\theta \$$

However I have no idea where to go from here? Have I calculated the limits wrong or am I missing something?

Thank you

• Ok, in order to integrate the $\theta$ function, go for integration by parts, choose $f=\theta$ and $g'=csc^2\theta$ – imranfat Mar 18 '16 at 23:56
• Wow, I can't believe I didn't see that! Thank you for guiding me! – THISISIT453 Mar 18 '16 at 23:58
• Are there any easier ways to achieve this answer? – THISISIT453 Mar 19 '16 at 0:00
• hmmm, I thought my approach was kind of easy, in the sense that behind the integral sign you get a $cot\theta$ and that has a standard anti derivative in the form of $ln|sin\theta|$ – imranfat Mar 19 '16 at 0:02
• The limits of $y$ should be from $x$ to $1$? – Mhenni Benghorbal Mar 19 '16 at 0:17

You need this: $$\int_0^{\pi/4}\int_0^{1/\cos\theta} \theta r\,dr\,d\theta.$$ Notice that $x$ goes from $0$ to $1$. So you have a right triangle whose "adjacent" side has length $1$, and $r$ is the length of the hypotenuse. $$\frac{\text{adjacent}}{\text{hypotenuse}} = \cos\theta = \frac 1 r,$$ so $r$ goes up to $1/\cos\theta$. So you get \begin{align} & \int_0^{\pi/4} \frac \theta {2\cos^2\theta} \,d\theta = \frac 1 2 \int_0^{\pi/4} \theta \Big( \sec^2\theta \,d\theta \Big) = \overbrace{\frac 1 2 \int \theta\,dv = \frac 1 2 \theta v - \frac 1 2 \int v\,d\theta}^\text{integration by parts} \\[12pt] = {} & \frac 1 2 \left[ \theta\tan\theta \vphantom{\frac 1 1} \right]_0^{\pi/4} - \frac 1 2 \int_0^{\pi/4} \tan\theta\,d\theta = \text{etc.} \end{align} Recall that $$\int\tan\theta\,d\theta = \int \frac{\sin\theta}{\cos\theta}\,d\theta = \int \frac {-1}{\cos\theta} \Big( - \sin\theta\,d\theta\Big) = \int \frac {-1} u\,du = \text{etc.}$$
• ok, You have $\displaystyle \int_x^1$ rather than $\displaystyle \int_0^x$. So $\theta$ would need to go from $\pi/4$ to $\pi/2$. $\qquad$ – Michael Hardy Mar 19 '16 at 0:19
In Cartesian, if your limits are set up correctly at the beginning, your region is $x=y, y=1, x=0$ And since $x<1$ you are going to get a negative value when you integrate.
After you have converted, you have $r = \csc \theta$ which is consistent with y=1. If the rest of the set up was right, you should have $\theta$ from $\pi/4$ to $\pi/2$
Alternatively, if your region was $x = 1, y=0, x=y$ your limit become $r$ from $0$ to $\sec\theta, \theta$ from $0$ to $\pi/4$