I have 2 questions I would like assistance with.

1) Find the area of the region bounded by the graphs $y=5x, y=15x, y=\frac{4}{x}, y=\frac{8}{x}$

This was very difficult and tedious.

I had trouble jotting it down so i drew it using a computer program, and if I understood the question correctly, we are required to find the area that i marked with a circle in this picture https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-prn2/v/t35.0-12/10360308_10152462327944628_1787541395_o.jpg?oh=8f0647c808a4e0cbfad86286492e792f&oe=53787F81&gda=1400376869_ad976f8063efd7f49d8eb9b1ee9a894d

I will spare you my entire calculations, it was easily over 3 pages long, basically since I have no easy way of finding this integral, i first found the area of small region yellow blue and green, then I drew a line paralel to x axis from the point of convergence of red and yellow, found the area of the region bound between red, blue, and the paralel line we drew.

Then I found the area of the entire triangle, and subtracted the areas we found before. that why i isolated the area requested and my final answer is $\frac{2\sqrt{30}}{15}-4\sqrt{12}+4\sqrt{60}-8-log(9) \approx 12.0549$

Now just multiply by 2 to get the answer for both circles.

There has to be an easier way. I'd like to compare answers with someone more knowledgable.

2) Find the area of the region bound inside the loop $(x+y)^3 =axy$ , $a >0$ in the first quadrant.

Hint: Use the transform $x=rcos^2 \theta$, $y=rsin^2 \theta$

What I did:

Basically I just want to make sure I got the limits right.

if we do the transform suggested, we get $r^3 =ar^2 cos^2\theta sin^2 \theta$

divide by $r^2$ to get $r=acos^2 \theta sin^2 \theta$ so $r$ varies from $0$ to $a\cos^2\theta sin^2 \theta$

How can we know how $\theta$ varies?

  • 1
    $\begingroup$ for the first problem, changes variables to $u=\frac{y}{x}$ and $v=xy$. $\endgroup$ – symplectomorphic May 16 '14 at 15:04
  • $\begingroup$ You called it "loop"...why? It certainly is a loop which is contained in the first quadrant, so $\;0\le\theta\le\frac\pi2\;$... $\endgroup$ – DonAntonio May 16 '14 at 15:04
  • $\begingroup$ I just tried doing the variable change you suggested, I got that the answer is $800$...doesnt seem right. The jacobian of the suggested transform is $2u$. the integral $\int_{4}^{8} \int_{5}^{15} 2ududv = 800$ $\endgroup$ – Oria Gruber May 16 '14 at 15:14
  • $\begingroup$ The Jacobian isn't $2u$. I've posted an answer. $\endgroup$ – symplectomorphic May 16 '14 at 15:42

For the first problem, change variables to $u=\frac{y}{x}$ and $v=xy$. Then you want to integrate over the region with $(u,v)\in[5,15]\times[4,8]$. By the Change of Variables Theorem, $$\int_{\phi(S)}f(x,y)\,dxdy=\int_Sf(u,v)\,|\det D\phi|\,dudv$$ where $D\phi$ is the Jacobi matrix of the mapping $$\phi(u,v)=\langle \underbrace{v^{1/2}u^{-1/2}}_{=x},\underbrace{v^{1/2}u^{1/2}}_{=y}\rangle$$

(Note that this will only yield the region in the first quadrant, because $x$ and $y$ are positive. To compute the inverse $\phi$, you have to separate the two regions, but they will have the same area by symmetry.) The Jacobi matrix is $$D\phi=\begin{pmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}v^{1/2}u^{-3/2} & \frac{1}{2}v^{-1/2}u^{-1/2} \\ \frac{1}{2}v^{1/2}u^{-1/2} & \frac{1}{2}u^{1/2}v^{-1/2}\end{pmatrix}$$ The determinant of this matrix is $$-\frac{1}{4}u^{-1}-\frac{1}{4}u^{-1}=-\frac{1}{2u}$$

Since $u>0$ in our region of integration, the absolute value of this determinant is $\frac{1}{2u}$. So our integral is $$\int_4^{8}\int_5^{15}\frac{1}{2u}\,dudv=\frac{1}{2}\cdot 4\ln3=2\ln 3=\ln 9$$

You have to double this if you want to include the region in the third quadrant.

  • $\begingroup$ But I thought since $u=\frac{y}{x}$ and $v=xy$ then the jacobian is $\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix}$ = $\begin{pmatrix} -\frac{y}{x^2} & \frac{1}{x} \\ y & x\end{pmatrix}$. The determinant of which is equal to $-2\frac{y}{x}=-2u$ or in absolute values, $2u$ $\endgroup$ – Oria Gruber May 16 '14 at 15:48
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    $\begingroup$ No, that's the wrong map. You need the Jacobi matrix of the map from $(u,v)$ to $(x,y)$, not the map from $(x,y)$ to $(u,v)$. In problems like these, the transformation is usually easiest to think of as a map from $(x,y)$ to $(u,v)$ (as I gave in my hint), but then you have to find the inverse mapping from $(u,v)$ to $(x,y)$ to do the integral. Or, if finding the inverse is too hard, you can just use the Inverse Function Theorem to conclude that the determinants of the two derivatives will be reciprocals. In this case, you want the reciprocal of $2u$. $\endgroup$ – symplectomorphic May 16 '14 at 15:53

Deepest apologies, I gave a mistaken statement on your first problem. The logs don't cancel.

The easy way to do this problem is to change variables to u = y/x and v = xy. Then each of the bounding curves becomes a horizontal or vertical straight line. Since you already know the vertex points, the equations of the new lines are trivial to get. The Jacobian of the transformation is -2u so you are integrating -1/(2u) over that rectangle. The last tricky part is that the identities of the vertical have swapped, so you have to negate the answer. (Thus the answer comes out positive.)

For the second problem, the polar coordinate equation is r = (a/4) sin^2(2 theta). In the first quadrant, theta ranges from 0 to pi/2, and that traces the curve once, so those are the right limits. By the way, the integral is easy if you transform with phi = 2 theta and notice that the integral from 0 to pi of sins^2 phi is the same as the integral from 0 to pi of cos^2 phi. Hint - add the two identical integrals.

  • $\begingroup$ The hyperbolas $y=\frac{4}{x}$ and $y=\frac{8}{x}$ are not confocal. Moreover, the correct answer does involves logs. $\endgroup$ – symplectomorphic May 16 '14 at 17:04

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