Can I double integrate a function through r and theta instead of x and y?

Question

Let's suppose a function $z=f(x, y)$ which has linear relationship with the distance from origin. For example we have a function $z=f(x, y)=max(-a \sqrt{x^2+y^2}+b,0)$, where $a,b>0$. And we're calculating it's double integral $\int_{\infty}^{-\infty} \int_{\infty}^{-\infty} f(x, y) \,dx \,dy$. Since the function involves max function, it would be easier to integrate it in polar coordinate form. So I'm wondering will the above double integral equals to $\int_{0}^{\frac{b}{a}} \int_{0}^{2\pi} -ar+b \,d\theta\ r \,dr$? Cause if both of them equals to the other, integrating through polar coordinate is definite integration compare to indefinite integration, which is way easier.

Answer 1

Short answer: The two integrals do equal, and are both definite as the bounds of the integration are given.

Long Answer: If we take your function to be, $$z=f(x,y)= \begin{cases} b-a\sqrt{x^2+y^2}, & \text{for}\sqrt{x^2+y^2}\le\frac{b}{a} \\ 0, & \text{for}\sqrt{x^2+y^2}\ge\frac{b}{a} \end{cases}=g(r,\theta)= \begin{cases} b-a r, & \text{for }r\le\frac{b}{a} \\ 0, & \text{for }r\ge\frac{b}{a} \end{cases}$$

Which 'breaks' you function into two regions, hence the integral can also be broken in to two regions. And if we take the

Then your integral, in x and y is $$\int_{\infty}^{-\infty} \int_{\infty}^{-\infty} f(x, y) \,dx \,dy= \iint_{\sqrt{x^2+y^2}\le\frac{b}{a}} b-a\sqrt{x^2+y^2} \,dx\,dy+ \iint_{\sqrt{x^2+y^2}\ge\frac{b}{a}} 0 \,dx\,dy $$ or in polars $$\int_{\infty}^{-\infty} \int_{\infty}^{-\infty} f(x, y) \,dx \,dy= \iint_{r\le\frac{b}{a}} b-ar \,dr\,rd\theta+ \iint_{r\ge\frac{b}{a}} 0 \,dr\,rd\theta$$

and as the integral of zero across any region is zero, thus the integral of zero disappear. Hence $$\int_{\infty}^{-\infty} \int_{\infty}^{-\infty} f(x, y) \,dx \,dy= \int_{0}^{\frac{b}{a}} \int_{0}^{2\pi} -ar+b \,d\theta \,rdr$$ Hopfully that helps.

Answer 2

Yes, you're quite right. Changing to polar coordinates does simplify the problem considerably, and I believe the expression you've obtained for the integral in polar coordinates is correct.