# How do I find the limits to this double integral and convert into polar?

With $$M = \{(x, y) ∈ R^2: 1 ≤ x^2 − y^2 ≤ 9, 2 ≤ xy ≤ 4, x, y ≥ 0\}$$, calculate $$\int_M (x^2+y^2) \mathrm dV$$

What I´ve managed so far is to convert $$x$$ and $$y$$ into polar coordinates with $$x=r\cos(\theta)$$ and $$y=r\sin(\theta)$$ and got $$\iint \left[(r\cos(\theta))^2+(r\sin(\theta))^2\right] r\mathrm dr\mathrm d\theta$$

is it right to use polar coordinates? if so, how do I set the limits of this double integral? Thx

• Sketch the region, find intersection points of curves in first octant so you know the bound area over which you have to integrate. Commented Nov 15, 2020 at 16:40
• Why polar? I would recommend trying $u=x^2_y^2$, $v=xy$. It works out wonderfully. Commented Nov 16, 2020 at 1:05
• Oops, that's $x^2-y^2$. Commented Nov 16, 2020 at 1:24

Our region (highlighted in yellow) is bound by curves $$x^2 - y^2 = 1, x^2 - y^2 = 9, xy = 2$$ and $$xy = 4$$ in the first octant as ($$x \geq 0, y \geq 0$$).

Equating $$x^2 - y^2 = 1, xy = 2$$,

$$(\frac{2}{y})^2 - y^2 = 1 \implies y^4 + y^2 = 4$$. Consider this as a quadratic in $$y^2$$ and solve for $$y^2$$. We find $$y \approx 1.25$$ and so $$x = \frac{2}{y} = \approx 1.6$$.

Similarly find intersection with $$xy = 4$$.

Then find intersection of $$x^2-y^2 = 9$$ with $$xy = 2$$ and $$xy = 4$$.

We obtain intersection points as $$(1.6, 1.25), (2.13, 1.88), (3.08, 0.65)$$ and $$(3.25, 1.23)$$.

Keeping things simple, I will set up the integral in $$3$$ parts (please see sketch) -

i) $$1.6 \leq x \leq 2.13$$ between curves $$xy \geq 2$$ and $$x^2-y^2 \geq 1$$

ii) $$2.13 \leq x \leq 3.08$$ and $$2 \leq xy \leq 4$$

iii) $$3.08 \leq x \leq 3.25$$ between $$x^2-y^2 \leq 9$$ and $$xy \leq 4$$.

$$\displaystyle I_1 = \int_{1.6}^{2.13}\int_{2/x}^{\sqrt{x^2-1}} (x^2+y^2) dy dx \approx 1.48$$

$$\displaystyle I_2 = \int_{2.13}^{3.08}\int_{2/x}^{4/x} (x^2+y^2) dy dx \approx 6.02$$

$$\displaystyle I_3 = \int_{3.08}^{3.25}\int_{\sqrt{x^2-9}}^{4/x} (x^2+y^2) dy dx \approx 0.5$$

$$I = I_1 + I_2 + I_3 \approx \fbox{8}$$

Double-check all my arithmetic. Even if I made no mistakes, it'll be instructive.

Since $$M$$ is defined by $$1\le r^2\cos2\theta\le9,\,4\le r^2\sin2\theta\le8,\,0\le\theta\le\pi/2$$, the integration range is$$0\le\theta\le\pi/4,\,\max\{\sec2\theta,\,4\csc2\theta\}\le r^2\le\min\{9\sec2\theta,\,8\csc2\theta\}.$$The constraint $$\theta\le\pi/4$$ follows from $$\cos2\theta\ge0$$. No, wait, we need to restrict $$\theta$$ further. Since $$\sec2\theta\le8\csc2\theta$$, $$\theta\le\tfrac12\arctan8$$. Similarly, $$\theta\ge\tfrac12\arctan\tfrac49$$. On this, we integrate $$r^3drd\theta$$, giving\begin{align}I&=\int_{\tfrac12\arctan\tfrac49}^{\tfrac12\arctan8}d\theta\int_{\sqrt{\max\{\sec2\theta,\,4\csc2\theta\}}}^{\sqrt{\min\{9\sec2\theta,\,8\csc2\theta\}}}r^3dr\\&=\tfrac14\int_{\tfrac12\arctan\tfrac49}^{\tfrac12\arctan8}d\theta\left[(\min\{9\sec2\theta,\,8\csc2\theta\})^2-(\max\{\sec2\theta,\,4\csc2\theta\})^2\right]\\&=\frac14\left(\int_{\tfrac12\arctan\tfrac49}^{\tfrac12\arctan\tfrac89}48\csc^22\theta d\theta+\int_{\tfrac12\arctan\tfrac89}^{\tfrac12\arctan4}(81\sec^22\theta-16\csc^22\theta)d\theta+\int_{\tfrac12\arctan4}^{\tfrac12\arctan8}80\sec^22\theta d\theta\right).\end{align}Using\begin{align}\int_{\tfrac12\arctan a}^{\tfrac12\arctan b}\csc^22\theta d\theta&=\left[-\frac12\cot2\theta\right]_{\tfrac12\arctan a}^{\tfrac12\arctan b}\\&=\frac{b-a}{2ab},\\\int_{\tfrac12\arctan a}^{\tfrac12\arctan b}\sec^22\theta d\theta&=\left[\frac12\tan2\theta\right]_{\tfrac12\arctan a}^{\tfrac12\arctan b}\\&=\frac{b-a}{2},\end{align}we can finish:$$4I=48\frac{\tfrac89-\tfrac49}{2\tfrac89\tfrac49}+81\frac{4-\tfrac89}{2}-16\frac{4-\tfrac89}{2\cdot4\cdot\tfrac89}+80\frac{8-4}{2}=27+126-7+160=306,$$making the original integral $$76\tfrac12$$.

enter image description hereProps to J.G. I also figured it out, hope it is correct

• Your bounds may not be correct. Please see the sketch in my answer. Commented Nov 15, 2020 at 21:41