Calculate the double integral: $\iint \frac{x^2}{x^2 + y^2}dxdy$ over the area $T = \{ (x, y) \in R^2 \mid 0 \le y \le x, \frac{1}{2} \le x \le 1\}$

Question

I first started by setting the bounds inside the integral as follows:

$$ T = \{ (x, y) \in R^2 \vert 0 \le y \le x, \frac{1}{2} \le x \le 1\}$$

$$\int_\frac{1}{2}^1 \int_0^x \frac{x^2}{x^2 + y^2}dydx$$

This integral is straight forward to evaluate:

\begin{align} \int_\frac{1}{2}^1 \int_0^x \frac{x^2}{x^2 + y^2}dydx & = \int_\frac{1}{2}^1 \int_0^x \frac{1}{1 + ({\frac{y}{x}})^2}dydx\\ & =\int_\frac{1}{2}^1 \left.\left[x\arctan(\frac{y}{x})\right]\right\vert_{y=0}^{y=x} dx\\ & =\int_\frac{1}{2}^1 x\arctan(1) dx\\ & =\frac{\pi}{4}\int_\frac{1}{2}^1 x dx\\ & =\frac{\pi}{4}\left.\left[\frac{x^2}{2}\right]\right\vert_\frac{1}{2}^1\\ & =\frac{\pi}{4}(\frac{1}{2}-\frac{1}{8})\\ & =\frac{3\pi}{32} \end{align}

However, the solution shows another answer:

Solution screenshot Sorry it's in french, but to translate: they consider the area $T \cup T'$ which is symmetric (since $T \cap T' = \emptyset$ we can use the additivity property). From there, their final answer is $\frac{3}{16}$.

To top it off, I checked the value of the integral with symbolab just in case I made a calculation error, and it gives me $\frac{3\pi}{32}$, which is what I got.

The answer in the solution makes sense, but I don't understand why my method is wrong. I figure my error lies in how I interpreted the integral bounds, but when drawing the area $T$ on the cartesian plane, the bounds seem perfectly fine. If someone could please clarify the error.

Answer 1

I believe your answer is correct and the solution is wrong. The very first equality in the solution is wrong:

$$ \iint_T \frac{x^2}{x^2+y^2} dx dy= \frac{1}{2}\iint_{T\cup T'} \frac{x^2}{x^2+y^2} dx dy = \dots $$

$T'$ may have the same area, but the integrand is not symmetric around the $y = x$ line, so the integral over $T$ is not half the integral over $T \cup T'$.

EDIT: In order for the solution to be justified, the function would have to look the same when reflected on the $y = x$ line. You can check that by swapping $x$ and $y$. If we do that on the integrand, $x^2 / (x^2 + y^2)$, we end up with $ y^2 / (x^2 + y^2)$: the denominator is symmetric under the interchange of $x$ and $y$, but the numerator is different, so the integrand is not symmetric under that interchange and so its graph looks different when reflected on the $y = x$ line.

Note by the way that "symmetric" does not mean anything by itself: you have to specify an operation under which it is symmetric, e.g reflection on a line. The above shows that the integrand is not symmetric when reflected on the $y = x$ line, but it remains unchanged if you change $x$ to $-x$, so its graph is symmetric when reflected on the y-axis (the $x=0$ line); it's also symmetric if you change $y$ to $-y$, so its graph is symmetric when reflected on the x-axis (the $y=0$ line). So if you know what it looks like in the first quadrant, then the reflection symmetries will allow you to know what it looks like in all four quadrants. So it has a number of symmetries, it just doesn't have the one that the "solution" wanted.

Answer 2

Note that

$$ \int_\frac{1}{2}^1 \int_0^x \frac{x^2}{x^2 + y^2}dydx = \int_0^\frac{1}{2} \int_\frac{1}{2}^1 \frac{x^2}{x^2 + y^2}dxdy+\int_\frac{1}{2}^1\int_y^1 \frac{x^2}{x^2 + y^2}dxdy $$.

and your answer is correct.