0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Sketch the region, find intersection points of curves in first octant so you know the bound area over which you have to integrate. $\endgroup$
    – Math Lover
    Commented Nov 15, 2020 at 16:40
  • $\begingroup$ Why polar? I would recommend trying $u=x^2_y^2$, $v=xy$. It works out wonderfully. $\endgroup$ Commented Nov 16, 2020 at 1:05
  • $\begingroup$ Oops, that's $x^2-y^2$. $\endgroup$ Commented Nov 16, 2020 at 1:24

3 Answers 3

1
$\begingroup$

enter image description here

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 $xy \leq 4$.

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

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

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

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

$\endgroup$
0
$\begingroup$

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 $76\tfrac12$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Your bounds may not be correct. Please see the sketch in my answer. $\endgroup$
    – Math Lover
    Commented Nov 15, 2020 at 21:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .