# Rewrite double integral in polar coordinates $\int_{0}^{3}\int_{x/3}^{1}fdydx$

I have the following... $$\int_{0}^{3}\int_{x/3}^{1}fdydx$$

I need to rewrite this in polar coordinates. I graphed the triangle and can see that...

$$\tan(\theta) = \frac{1}{3}$$

But I do not know how to use this information to rewrite my integral into polar coordinates using $\pi$.

• What is meant by f? – SchrodingersCat Oct 5 '17 at 5:22
• @SchrodingersCat it is just a function, it is not necessary for it to be defined – Bolboa Oct 5 '17 at 5:24
• @SchrodingersCat it is $f(x,y)$, I need only change the bounds and to rewrite $f(x,y)$ as $rdrd\theta$ – Bolboa Oct 5 '17 at 5:26
• Hint: Solve for $\theta$. Then use x = r cos ($\theta$) and y = r sin ($\theta$) and you should be able to rewrite the integral. – Mathemagician1234 Oct 5 '17 at 5:27
• @Bolboa Okay... – SchrodingersCat Oct 5 '17 at 5:27

For simplicity, I will assume that $f(x,y)= 1$.

Note that if you are given $f(x,y)$ explicitly, you have to re-express $f(x,y)$ in terms of $r$ and $\theta$.

Using the hint that $x = r\cos\theta$ as mentioned in the comment section, and $y=1$,

$$1=r\sin \theta \implies r = \frac{1}{\sin \theta}$$

Also,

$$\tan(\theta ) = \frac{1}{3} \implies \theta = \arctan(\frac{1}{3})$$

So we can setup the double integral as the following,

$$\int_{\arctan\bigg(\dfrac{1}{3}\bigg)}^{\dfrac{\pi}{2}}\int_{0}^{\dfrac{1}{\sin \theta}}r \text{ }dr d\theta$$

Below is a (ugly drawn) picture for a fixed $(r,\theta)$.

(Another exercise) Rewrite the double integral in polar coordinates

$$\int_{0}^{3} \int_{0}^{x/3}\text{ } dydx$$

Let $x=3$,

$$3 = r\cos \theta \implies r = \dfrac{3}{cos\theta}$$

Also, $$\tan(\theta ) = \frac{1}{3} \implies \theta = \arctan(\frac{1}{3})$$

So we can setup the double integral as the following,

$$\int_{0}^{\arctan\bigg(\dfrac{1}{3}\bigg)}\int_{0}^{\dfrac{3}{\cos \theta}}r \text{ }dr d\theta$$

Below is another (ugly drawn) picture for a fixed $(r,\theta)$.

$$\int ^{\frac {\pi}{2}} _{\arctan \frac {1}{3}} \int ^{\frac {1}{\sin \theta}} _{0} f(r\cos \theta, r \sin \theta)r dr d \theta$$ I think the area should be the upper triangle but not the lower.