# Double integral new coordinate system calculation

1. Apr 2, 2013

### cambo86

1. The problem statement, all variables and given/known data
This is a 2 part question. I'm fine with the first part but the 2nd part I'm struggling with.

The first part asks us to calculate the double integral,
$\int\int$Dx2dA
for, D = {(x,y)|0≤ x ≤1, x≤ y ≤1}
For this part I got an answer of 1/4.

For the 2nd part we introduce a new coordinate system for D,
x = 1-st, 0≤ s ≤1
y = s, 0≤ t ≤1

3. The attempt at a solution
$\int\int$Dx2dA
= $\int_0^1\int_0^1$(1-st)2dt ds
= $\int_0^1\int_0^1$(1-2st+s2t2)dt ds
= $\int_0^1$t-st2+$\frac{1}{3}$s2t3 ds, from t=0 to t=1
= $\int_0^1$1-s+$\frac{1}{3}$s2 ds
= s-$\frac{1}{2}$s2+$\frac{1}{9}$s3, from s=0 to s=1
= 1-$\frac{1}{2}$+$\frac{1}{9}$
= $\frac{11}{18}$

I feel like,
dA $\neq$ dt ds
I'm not sure what it equals though. I thought I could use polar coordinates but I don't have a constant radius.

2. Apr 2, 2013

### HallsofIvy

Staff Emeritus
Okay, so you got this wrong. In more detail this is
$$\int_{x= 0}^1\int_{y= x}^1 x^2dydx= \int_0^1 (1- x)x^2 dx$$
What do you get for that?

You can't just replace "dydx" with "dtds". Since you are now measuring distance differently, the area will be measured differently. Use the Jacobian:
$$\left|\begin{array}{cc}\frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t}\end{array}\right|$$$$= \left|\begin{array}{cc}-s & -t \\ 0 & 1 \end{array}\right|= -s$$

So you have to replace "dxdt" with -s dsdt as well as replacing $x^2$ with $(1- st)^2$.

Yes, that is the difficulty. Use the Jacobian, as I said before.

3. Apr 2, 2013

### cambo86

Sorry, the lower bound on y for the first part should have been 1-x but I should be able to work through the rest of the reply and get an answer. Thank you for the detailed reply.