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I'm trying to evaluate:

$$\iint_D \left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy$$

where $D_{xy}$ is the disk $x^2+y^2\le a^2$.

The exercise is to use change of variables to solve this integral.

My solution

I chose $\varphi (r,\theta)=(ra\cos\theta,ra\sin\theta)$, where $0\le r\le 1$ and $0\le \theta\le 2\pi$ to be the change of variables.

The determinant of the Jacobian is $ra^2$ and \begin{align*} &\iint_{D_{xy}}\left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy \\&=\int_0^{2\pi}\int^1_0\left(\sqrt{a^2-r^2a^2}-ra\right)ra^2 drd\theta\\ &=2\pi a^3\int^1_0 \left(r\sqrt{1-r^2}-r^2 \right)dr\\ &=2\pi a^3\left(\int^1_0r\sqrt{1-r^2}dr-\int^1_0r^2dr\right)\\ &=2\pi a^3\left( \frac{1}{3}-\frac{1}{3} \right)\\ &=0 \end{align*}

I would like to know where I'm mistaken. The answer in the end of the book shows $\pi a^3/3$.

erfink
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user42912
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2 Answers2

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The answer is zero; the answer in the book is incorrect.

To verify, let's show that the following integrals (from your derivation above) are equal:

$$ A = \int_0^1 r\sqrt{1 - r^2} dr$$ $$ B = \int_0^1 r^2 dr $$

Let's change variable in $A$:

$$\quad t^2 = 1 - r^2 \Rightarrow 2tdt = -2rdr\Rightarrow rdr = -tdt.$$

Hence,

$$ A = \int_0^1 \sqrt{1 - r^2}\;rdr = - \int_1^0 t^2 dt = \int_0^1 t^2 dt$$

Thus, $A$ and $B$ are equal.

Hoc Ngo
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Just for fun, I figured I'd show that the answer is indeed zero by a direct substitution without any evaluation of an integral.

Let $$ (x,y)=\frac{\sqrt{a^2-\hat{x}^2-\hat{y}^2}}{\sqrt{\hat{x}^2+\hat{y}^2}} (\hat x,\hat y) $$ Now, a quick calculation will confirm that the determinant of the Jacobian is $-1$ (and so the multiplying factor is 1), and that it maps the disc of radius $a$ onto itself.

Applying our transformation, we get $$\begin{align} I&=\iint_D\left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}\right)dxdy\\ &=\iint_D\left(\sqrt{\hat x^2+\hat y^2}-\sqrt{a^2-\hat x^2-\hat y^2}\right)d\hat xd\hat y\\ &=-I \end{align}$$ Therefore, $I=0$.

Glen O
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