# Area is symmetric

1. Jan 30, 2007

### bodensee9

1. The problem statement, all variables and given/known data
Help!! I was wondering if anyone can help me integrate:

∫ x^2tan x + y^3 +4 dA, where D is the region represented by D = {(x,y)|x2+y2≤2}

2. Relevant equations
I think that the area is symmetric, and so basically you only need to evalute from 0≤ x ≤ √(2-y^2) and 0≤ y ≤ √2. Or you can do x first and evaluate it from 0 ≤ y ≤ √ (1-x^2) and 0≤ x ≤ √2. But I'm not sure how to evaluate x^2 tan x? I don't think doing dy first will help either since I get the following: x^2*√ (1-x^2)*tan x? Thanks!!!

2. Jan 30, 2007

### Tom Mattson

Staff Emeritus
I've been playing around with this, and I think I've made some progress.

Integrate the first term in the integrand with respect to $x$ first. Do it by parts with $u=x \tan(x)$ and $dv=x dx$. You should be able to express the integral in terms of $\int x^3\sec^2(x)dx$. Integrate that by parts with $u=x^3$ and $dv=sec^2(x)dx$.

Give that a try and see how it goes. I haven't finished it yet, but it looks like it will work.

3. Jan 31, 2007

### HallsofIvy

Staff Emeritus
This is, of course, the same as
$$\int x^2 tan x dA+ \int y^3 dA+ \int 4 dA$$
Yes, the region $D= {(x,y)| x^2+ y^2\le 2}$ is a circle and so is symmetric. But why do you then say you need only integrate in the first quadrant? x2tan x and y3 are both ODD functions. Their integrals on opposite sides of the axes will cancel, not add. It looks to me like this is just $\int 4 dA$ or just 4 times the area of the circle.

4. Jan 31, 2007

### bodensee9

Oh I see!! Thanks!!!