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 $\pi$.
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$\begingroup$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.
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