Evaluate the definite integral $$\iint_{D} 2x-y \,dA$$ where $D$ is bounded by the circle with center at the origin and a radius $2$.
This particular problem looks like a simple case of converting a definite integral to polar coordinates then solving.
I know that in polar coordinates:
- $x$ becomes $rcos(\theta)$
- $y$ becomes $rsin(\theta)$
- $dA$ becomes $r\, dr\, d\theta$
The integral is bounded by the radius, $0 \le r \le 2$, and $0 \le \theta \le 2\pi$. So I have the integral:
$$\int_0^{2\pi} \int_0^2 2(rcos(\theta))-(rsin(\theta)) \, r \, dr \,d\theta$$
$$\int_0^{2\pi} \int_0^2 r^2(2cos(\theta)-sin(\theta)) \, dr \,d\theta$$
$$\frac{2^3}{3} \int_0^{2\pi} 2cos(\theta)-sin(\theta) \,d\theta$$
But the remaining integral $\int_0^{2\pi} 2cos(\theta)-sin(\theta) \,d\theta$ evaluates to $0$. Intuitively, a circle with some radius cannot have an area that is $0$, so this result is obviously incorrect.
The visual representation of the inner integral is as follows:
Is this answer correct? Where did I go wrong?
Thanks so much for your help!
Best Answer
You're not wrong. The answer is 0. This is because the process of doing a double integral means finding the volume. When the answer is 0, this means no depth exists for the integral, so you are left with the xy plane.