# Double integrals in polar coordinates?

So, the question asks me to find the result of this double integral over $S: \iint(2e^{x^2}) dA$ where $S$ is a triangle in the (x,y) plane bounded by the x-axis and the lines $y=x$ and $x=1$. Now, I'm very new to double integrals with polar coordinates. With the conventional way (without using polar coordinates), I have found that the answer is $\frac 23$. Can you tell me how to solve this? Thank you so much!

• Hint: try writing the bounds $y=x$, $x=1$, and the $x$-axis in polar coordinates. – kccu May 16 '17 at 18:10
• The bound in my conventional integral is y from y to 1 and x from 0 to 1, where I finally found 2/3. First, I'm not sure, are those bounds right? Second, how to translate the bounds into the polar coordinates? – Alexander May 16 '17 at 18:14
• Do you know how to write $x$ and $y$ in terms of polar coordinates? Try substituting those into the equations $y=x$, $x=1$, and the equation for the $x$-axis. – kccu May 16 '17 at 18:23

You can visualize the region $S$ like so:

You can integrate first along $y$ from $y = 0$ to $y = x$, and then integrate that along $x$ from $x = 0$ to $x = 1$: $$\iint_S 2e^{x^2} \, dA = \int_0^1 \int_0^x 2e^{x^2} \, dy \, dx$$

You could also integrate by $x$ first, from $x = y$ to $x = 1$, and then $y = 0$ to $y = 1$, but the integral does not simplify when integrating by $x$ first.

So: \begin{align*} \iint_S 2e^{x^2} \, dA &= \int_0^1 \int_0^x 2e^{x^2} \, dy \, dx \\ &= \int_0^1 2xe^{x^2} \, dx \\ &= \left[e^{x^2} \right]_0^1 \\ &= e - 1 \end{align*}

Doing this integral in polar coordinates, we see that the angle from the $x$-axis goes from $\theta = 0$ to $\theta = \pi/4$. The distance from the origin, as we go along our values of $\theta$, go from $r = 0$ to the value of $r$ where $x = 1$, that is: $$r \cos \theta = 1 \quad\Longleftrightarrow\quad r = \sec \theta$$ Making the change of coordinate systems, we remember to note that: $$dA = dx\,dy = r \, dr\,d\theta$$ so our integral simplifies to: \begin{align*} \iint_S 2e^{x^2} \, dA &= \int_0^{\pi/4} \int_0^{\sec \theta} 2r e^{r^2 \cos^2 \theta} \, dr \, d\theta \\ &= \int_0^{\pi/4} \left[ \frac{1}{\cos^2 \theta} e^{r^2 \cos^2 \theta} \right]_{r = 0}^{\sec \theta} \, d\theta \\ &= \int_0^{\pi/4} \left[ \frac{1}{\cos^2 \theta} e^{r^2 \cos^2 \theta} \right]_{r = 0}^{\sec \theta} \, d\theta \\ &= \int_0^{\pi/4} (e-1)\sec^2 \theta \, d\theta \\ &= (e-1) \cdot \int_0^{\pi/4} \sec^2 \theta \, d\theta \\ &= (e-1) \cdot \left[ \tan \theta \right]_0^{\pi/4} \\ &= e-1 \end{align*}

Hints:

Notice that as $(x,y)$ spans the triangle, the angular polar coordinate $\phi$ spans the interval $[0,\pi /4]$. When $\phi = 0$ you are on the x axis and when $\phi = \pi/4$ you are just above the line $y = x$

Now let's take a look at the span of the radial coordinate $\rho$. It goes from zero (the triangle touches the origin) to the line $x = 1$. Remember that $x = \rho cos(\phi)$, then $0 \leq \rho \leq (cos(\phi))^{-1}$.

Hint:

The integral is equal to $$\iint_A2\mathrm e^{r^2\cos^2\theta}r\mathrm dr\,\mathrm d\theta.$$ Note the polar equation of the vertical leg of the triangle is $r\cos\theta=1\;(0\le \theta\le \pi/4)$.

Use Fubini, integrating first w.r.t. $r$.