# Thread: [SOLVED] A couple of double integrals

1. ## [SOLVED] A couple of double integrals

EDIT: Solved all of them.

I guess I don't know how to solve any of the following, because I keep getting the wrong answer all the time. It's mostly the boundaries that I'm having trouble with.

1) $\int{\int_D{x^3}}dxdy$ where $D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}$

2) $\int{\int_D{x}}dxdy$ where $D= \left\{(x,y)\in R^2; 1\leq x^2 + 2xy + 4y^2 \leq 1; x \geq 0; y \geq 0\right\}$

3) $\int{\int_D{\frac{dxdy}{(1+x^2-y^2)^2}}}dxdy$ where $D$ is the triangle with corners in $(0,0), (1, -1)$ and $(3,1)$.

So here's what I tried to do:

1)

$\left\{\begin{array}{l}u= x\\v = 3y\end{array}\right.$

$|J(u,v)| = \frac{1}{3}$

$1 \leq u^2 + v^2 \leq 9, u \geq v$

$\left\{\begin{array}{c}u= rcos\varphi\\v = rsin\varphi\end{array}\right.$

$|J(r, \varphi)| = r$

So the limits become $1 < r < 3$ and $\frac{-3\pi}{4}< \varphi < \frac{\pi}{4}$

2. Originally Posted by Spec
I guess I don't know how to solve any of the following, because I keep getting the wrong answer all the time. It's mostly the boundaries that I'm having trouble with.

1) $\int{\int_D{x^3}}dxdy$ where $D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}$

[snip]

So here's what I tried to do:

1)

$\left\{\begin{array}{l}u= x\\v = \frac{y}{3}\end{array}\right.$

$|J(x,y)| = \frac{1}{3}$

Boundaries?

$1 \leq u^2 + v^2 \leq 9$

$\left\{\begin{array}{c}u= rcos\varphi\\v = rsin\varphi\end{array}\right.$

$|J(r, \varphi)| = r$

Boundaries?

$\int{\int_D{r^3cos^3\varphi}}\cdot |J(u,v)| dxdy = \int{\int_D{\frac{r^4cos^3\varphi}{3}}}dxdy$
[snip]
I have time for one:

After your transformation to uv-coordinates (which was a good move by the way), you have $\frac{1}{3} \int \int_{D'} u^3 \, du \, dv$ where $D' = \left\{(u ,v) \in R^2; 1 \leq u^2 + v^2 \leq 9; u \geq 9v \Rightarrow v \leq \frac{u}{9}\right\}$.

Drawing a diagram that shows D' will be a big help for you.

Switch to polar coordinates and you have $\frac{1}{3} \int_{r = 1}^{r = 3} \int_{\varphi = \alpha}^{\varphi = \beta} r^4 \cos^3\varphi \, d\varphi \, dr$ where $\alpha = \tan^{-1} \left( \frac{1}{9} \right) - \pi$ and $\beta = \tan^{-1} \left( \frac{1}{9} \right)$.

3. I'm not getting the right answer by using that. It's supposed to be $\frac{121\sqrt{2}}{9}$, but I'm getting something way more complicated.

I would appreciate some help with the other integrals as well.

4. Originally Posted by Spec
I'm not getting the right answer by using that. It's supposed to be $\frac{121\sqrt{2}}{9}$, but I'm getting something way more complicated.

I would appreciate some help with the other integrals as well.
There's an error with the Jacobian. It's meant to be $\left| \begin{array}{cc}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right|$
$= \left| \begin{array}{cc}
1 & 0 \\
0 & 3 \end{array} \right| = 3$
.

I can't see any other error. But I get $\frac{5929 \sqrt{82}}{1681}$.

Maybe someone else can see what I can't (if there's anything to see).

If I have time I'll take a look at the others. I would have thought that the second one is similar to the first in many ways ......

5. Originally Posted by Spec
[snip]

3) $\int{\int_D{\frac{dxdy}{(1+x^2-y^2)^2}}}dxdy$ where $D$ is the triangle with corners in $(0,0), (1, -1)$ and $(3,1)$.

[snip]
The triangular region D is bounded by the lines $y = -x, ~ y = \frac{x}{3}\,$ and $y = x - 2$.

Note that $1 + x^2 - y^2 = 1 + (x - y)(x + y)$. This suggests making the transformation

u = x - y .... (1)

v = x + y .... (2)

(1) + (2): $x = \frac{u+v}{2}$.

(2) - (1): $y = \frac{-u+v}{2}$.

The Jacobian of the transformation is therefore $J(u, v) = \left| \begin{array}{cc}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
& \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right|$
$= \left| \begin{array}{cc}
\frac{1}{2} & \frac{1}{2} \\
& \\
-\frac{1}{2} & \frac{1}{2} \end{array} \right| = \frac{1}{2}$
.

The triangular region D transforms to a simple triangular region D' bounded by the following lines:

$y = - x \Rightarrow x + y = 0 \Rightarrow {\color{red}v = 0}$.

$y = x - 2 \Rightarrow x - y = 2 \Rightarrow {\color{red}u = 2}$.

$y = \frac{x}{3} \Rightarrow \frac{-u + v}{2} = \frac{u + v}{6} \Rightarrow {\color{red}v = 2u}$.

So the integral becomes $\frac{1}{2} \int_{u=0}^{u=2} \int_{v = 0}^{v = 2u} \frac{1}{(1 + uv)^2} \, dv \, du$ and it should be blue sky from here.

6. Originally Posted by mr fantastic
$\alpha = \tan^{-1} \left( \frac{1}{9} \right) - \pi$
$\beta = \tan^{-1} \left( \frac{1}{9} \right)$
Why are they $\tan^{-1} \left( \frac{1}{9} \right)$ ?

7. Originally Posted by wingless
Why are they $\tan^{-1} \left( \frac{1}{9} \right)$ ?

In transforming from xy-coordinates to uv-coordinates, the line x = 3y becomes u = 3(3v) = 9v => v = u/9, which has a gradient of 1/9 ......

The transformation is done to facilitate a switch to polar coordinates.

8. Originally Posted by Spec
I'm not getting the right answer by using that. It's supposed to be $\frac{121\sqrt{2}}{9}$, but I'm getting something way more complicated.

I would appreciate some help with the other integrals as well.

I got $\frac{121\sqrt{2}}{9}$. Here's my approach:

$\int{\int_D{x^3}}dxdy$ where $D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}$

Let $x = r \cos \theta$ and $y = \frac{1}{3}r \sin \theta$. Then
$D'=1\leq r^2 \leq 9;~\cos\theta \geq \sin\theta$.

$\int\int_D x^3~dx~dy = \int\int_{D'}r^3\cos^3 \theta \frac{r}{3}~dr~d\theta$.

From $1\leq r^2 \leq 9,~1\leq r \leq 3$.
From $\cos\theta \geq \sin\theta,~-\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}$.

$\int^{\frac{\pi}{4}}_{-\frac{3\pi}{4}}\int_1^ 3\frac{1}{3}r^4\cos^3 \theta ~dr~d\theta = \frac{121 \sqrt{2}}{9}$.

9. Originally Posted by wingless
I got $\frac{121\sqrt{2}}{9}$. Here's my approach:

$\int{\int_D{x^3}}dxdy$ where $D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}$

Let $x = r \cos \theta$ and $y = \frac{1}{3}r \sin \theta$. Then
$D'=1\leq r^2 \leq 9;~\cos\theta \geq \sin\theta$.

$\int\int_D x^3~dx~dy = \int\int_{D'}r^3\cos^3 \theta \frac{r}{3}~dr~d\theta$.

From $1\leq r^2 \leq 9,~1\leq r \leq 3$.
From $\cos\theta \geq \sin\theta,~-\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}$.

$\int^{\frac{\pi}{4}}_{-\frac{3\pi}{4}}\int_1^ 3\frac{1}{3}r^4\cos^3 \theta ~dr~d\theta = \frac{121 \sqrt{2}}{9}$.
The one thing I didn't check - Spec's transformation from xy-coordinates to uv-coordinates. I used the original transformation (see the quote in post #2) without checking it

Looking for my mistake, I see that it should have been v = 3y, NOT v = y/3. Then I notice Spec has edited the question to reflect this - s/he obviously realised the same thing (*ahem* it's good form to include the reason for an edit, by the way .....)

My mistake for not checking right from the start I knew it would be something simple.

Moral of the lesson - don't assume that even the simplest things are done correctly.

10. Originally Posted by mr fantastic
Moral of the lesson - don't assume that even the simplest things are done correctly.
That's so true