Let $R$ be the triangle defined by $−x\tan(\theta) \le y \le x\tan(\theta)$ and $x \le 1$ where theta is an acute angle. Sketch the triangle and calculate

\begin{equation*} \iint_R(x^2+y^2)\mathrm dA \end{equation*}

using polar coordinates.

I got $r=\frac{1}{\cos(\theta)}=\sec(\theta)$ but I am stuck on how to get the angle to solve this question... :/


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  • $\begingroup$ Can you please clarify? What is $-x\tan{\theta}$? $\endgroup$ – Sufyan Naeem May 19 '15 at 9:32
  • $\begingroup$ what do you mean?? y≤xtan(θ) and y>=-xtan(θ) $\endgroup$ – MathNoob May 19 '15 at 9:43
  • $\begingroup$ Don't quite understand which angle are you looking for.... $\endgroup$ – user99914 May 19 '15 at 9:48
  • $\begingroup$ there's only 1 angle which is theta for y=xtan(θ) $\endgroup$ – MathNoob May 19 '15 at 9:49
  • $\begingroup$ or the region where theta applies in this case $\endgroup$ – MathNoob May 19 '15 at 9:57

The triangle's angles are located at $(0,0)$, $(1,\tan\theta)$ and $(-1,\tan\theta)$. Using polar coordinates we have to integrate for a given angle $\theta'$ from the radius $0$ to $r=\sqrt{1^2+\tan^2\theta}=\sec\theta$. This gives $$I=\iint_R(x^2+y^2)\mathrm dA=\int_{-\theta}^{\theta}\mathrm d\theta'\int_0^{\sec\theta'}r^2\,r\mathrm dr=\frac14\int_{-\theta}^\theta\sec^4\theta'\,\mathrm d\theta'.$$ The result of this integral is, setting $t=\tan\theta'$ $$I=\frac14\;2\times\int_0^{\tan\theta}(1+t^2)\mathrm dt=\frac12\left(\tan\theta+\frac13\tan^3\theta\right).$$

Note added in edit (cartesian coordinates) $$\begin{split} I&=\int_0^1\mathrm dx\int_{-x\tan\theta}^{x\tan\theta}(x^2+y^2)\mathrm dy\\ &=\int_0^1\mathrm dx\left[x^2y+\frac13y^3\right]_{y=-x\tan\theta}^{y=x\tan\theta}\\ &=2\int_0^1\mathrm dx\left(x^3\tan\theta+\frac13x^3\tan^3\theta\right)\\ &=2\left(\tan\theta+\frac13\tan^3\theta\right)\int_0^1x^3\mathrm dx\\ &=2\left(\tan\theta+\frac13\tan^3\theta\right)\left[\frac{x^4}4\right]_{0}^1\\&=\frac12\left(\tan\theta+\frac13\tan^3\theta\right).\end{split}$$

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  • $\begingroup$ I dont really get the second section how you get tanθ and 0 on the limits/range (that thing on the integral sign) and how its (1+t^2) not (1+t^2)^2 $\endgroup$ – MathNoob May 19 '15 at 10:54
  • $\begingroup$ 1) The range: the function is even so $\int_{-\theta}^\theta=2\int_0^\theta$. 2) As $t=\tan\theta'$ we have $\mathrm dt=\sec^2\theta'\,\mathrm d\theta'$. Use $\sec^2\theta'=1+t^2$ to get the expression. $\endgroup$ – Tom-Tom May 19 '15 at 11:37
  • $\begingroup$ sorry to keep bothering you but how does 2∫θ 0 turn into 2∫tanθ 0 $\endgroup$ – MathNoob May 21 '15 at 7:42
  • $\begingroup$ The change of variable $t=\tan \theta$ modifies the boundaries of the integral. You should probably learn more about it. It's called integration by substitution. $\endgroup$ – Tom-Tom May 21 '15 at 7:53
  • $\begingroup$ i completely forgot about that... i feel so stupid now... anyway thanks $\endgroup$ – MathNoob May 21 '15 at 7:58

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