Interpreting and evaluating $\iint x^2 (1-x^2-y^2)dxdy $

Question

I am having trouble understanding what this question means:

Evaluate the integral

$$\iint x^2 (1-x^2-y^2)\,dxdy $$

over the interior of the circle of radius 1 centred on the origin.

First of all, when I am trying to understand the equation inside the integral sign, I see that $ (1-x^2-y^2)$ is the equation of a circle radius 1, but with $x^2 (1-x^2-y^2)$ it is not a circle, perhaps a figure of 8 type object. Therefore I am not sure what the question is meaning by "over the interior of the circle of radius 1".

I have tried simply plugging through the equations and I tried two different approaches:

• going into polar coordinates, which gave me an answer of $\pi /6$ when evaluating

$$\int_0^1\int_0^{2\pi} r^2cos^2 (\theta)(1-r^2)\;r\;d\theta dr $$

• considering a circle in cartesian coordinates gave me an answer of 4/45 which I'm pretty sure is wrong, from evaluating

$$\int_{-1}^1\int_0^1 x^2 (1-x^2-y^2)\;dydx$$

I think the first approach is correct, but it's difficult to know what I am doing wrong when I don't know what the question is referring to!

Answer 1

It seems to me that your trouble is in understanding the "over the interior of the circle of radius 1" (lets call it $R$) part, and what it has to do with $f(x,y)=x^2(1−x^2−y^2)$, your integrand.

Answer: It has nothing to do!

It is a function that goes under the integral sign, not an equation. With the notation we've established, your problem can be alternatively stated: Calculate $$\iint_Rf(x,y)dxdy,$$ to emphasise the distinction.

Geometrically, this integral evaluates the volume of the solid that is the intersection of the region between the graph of $f$ and the the $xy$-plane, and the cilinder $R\times \mathbb{R}$ (here $R$ is in the $xy$-plane and the $\mathbb{R}$ corresponds to the $z$-axis).

The error in your second approach is in the integration bounds. In the way you did it, you are integrating $f$ over the interior of the rectangle $[0,1]\times[-1,1]$. What you shoud do, instead is parametrize $R$ as $$R=\{(x,y)\in\mathbb{R}^2\ |\ -1<x<1\ ,\ -\sqrt{1-x^2}<y<\sqrt{1-x^2}\},$$ and calculate, (using Fubini's theorem) $$\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}\int_{-1}^1x^2(1−x^2−y^2)dxdy.$$

I think it will be much more simple to pass to polar coordinates, though, as you did. I did not check your result, but it sems right. Note that the integration bounds you used there corresponds to the rectangle $[0,1]\times[0,2\pi]$, that covers $R$ when you apply the polar coordinates transformation.