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I have the following double integral and the following domain.

$$\iint x^2 \tan(x) + y^3+ 4 dx dy$$ $$D=\{(x,y): x^2+y^2\le2\}$$

I know the domain is a circle and it can be described as:

$$D=\left\{(x,y): -\sqrt{2} \le x \le \sqrt{2} \text{ and } -\sqrt{2-x^2} \le y \le \sqrt{2-x^2}\right\}$$

I have tried integrating using this domain, but I have no idea how to handle $\int x^2\tan(x)$

I also tried to use polar coordinates, but $x^2\tan(x)$ also is something that I do not know how to handle in polar coordinates.

Can someone please help me.

Gary
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user3347814
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2 Answers2

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Clearly $$ \iint_D4dxdy=8\pi.$$ Let $$L=-x^2y\tan x, M=xy^3.$$ Then by Green Theorem, $$\iint_Dx^2 \tan(x) + y^3dx dy=\oint_\gamma Ldx+Mdy. $$ It is easy to get $$ \oint Ldx=\oint Mdy=0 $$ by symmetry. I omit the detail. So $$\iint x^2 \tan(x) + y^3+ 4 dx dy=8\pi.$$

xpaul
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Assuming $\operatorname{tg}(x)\equiv\tan x$, convert to polar coordinates and evaluate the trivial parts to get

$$\begin{align*} & \iint_\limits{x^2+y^2\le2} \left(x^2\tan x + y^3 + 4\right) \, dx \, dy \\ &= \int_0^{2\pi} \int_0^{\sqrt2} \left(r^3 \cos^2\theta \tan\left(r \cos\theta\right) + r^4 \sin^3\theta + 4r\right) \, dr \, d\theta \\ &= 8\pi + \int_0^{2\pi} \int_0^{\sqrt2} \cos^2\theta \tan\left(r \cos\theta\right) \, dr \, d\theta \\ &= \boxed{8\pi} - \int_0^{2\pi} \cos\theta \ln \cos\left(\sqrt2\,\cos\theta\right) \, d\theta \end{align*}$$

where in the last step, we use the elementary antiderivative,

$$\int \tan(r\cos\theta) \, dr = - \frac{\ln \cos\left(r\cos\theta\right)}{\cos\theta}$$

The remaining integral vanishes due to symmetry:

$$\begin{align*} & \int_0^{2\pi} \cos\theta \ln \cos\left(\sqrt2\,\cos\theta\right) \, d\theta \\ &= 2 \int_0^\pi \cos\theta \ln \cos\left(\sqrt2\,\cos\theta\right) \, d\theta \\ &= -2 \int_{-\tfrac\pi2}^{\tfrac\pi2} \sin\theta \ln \cos\left(\sqrt2\,\sin\theta\right) \, d\theta = 0 \end{align*}$$

user170231
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  • Thank you for your answer. Is the integral of tan(r cos(theta)) dr that you used and commun integral that I should know? Is there a way that I could get to this result by doing some substitutions or using other methods of integration? – user3347814 Sep 08 '23 at 13:21
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    You're welcome! There's nothing particularly special about that antiderivative since $\cos\theta$ is constant with respect to $r$, we are just integrating $\tan$ with a scaled argument. I just wanted to make it clear how the double integral is reduced to a single. – user170231 Sep 08 '23 at 14:14