# Double Integral and Polar, Really Need Help in the next few hours

1. Nov 20, 2012

### BenMcC

I have this problem and I cannot even begin to start it. I have to hand it in today in a few hours, and I have been stuck on it for what seems like for ever. It reads:

By using polar coordinates evaluate:

∫ ∫ (2+(x^2)+(y^2))dxdy
R

where R={x,y}:(x^2)+(y^2)≤4,x≥0,y≥0} Hint: The region R is the quarter of a circular disc with radius r=2 that lies in the fourth quadrant.

Any help would be greatly appreciated, I have no idea what else to do. Thanks

2. Nov 20, 2012

### arildno

What are the limits on the radius and the angular variable? (By what you have written, this lies in the first quadrant, not the fourth)

3. Nov 20, 2012

### BenMcC

It only states that the radius, r=2. And since it's in the fourth quadrant, the limits of the angle would be just 3∏/2 to 2∏ I believe. But I have no idea how to set up this problem

4. Nov 20, 2012

### arildno

Why not?
1. What is the area element in polar coordinates?
2. How is the integrand to be represented in polar coordinates?

5. Nov 20, 2012

### BenMcC

I typed the entire problem as it is on the assignment. There's nothing else with this problem

6. Nov 20, 2012

### arildno

Yes.
And you ought to know, by what you have learnt, how to:
1. Replace the quadratic area element with the proper polar coordinate area element.
2. Convert the integrand into a function of the polar coordinates.
3. How to set up the new limits of the integral.

7. Nov 20, 2012

### BenMcC

What I tried with this problem:
R:x≥0,y≥0
R≤4, so x^2+y^2≤4
-2≤x≤2, -sqrt(4-x^2)≤y≤sqrt(4-x^2)

The integral looked like
2∏
∫ ∫ (2+r^2)dr dθ
3∏/2 R

My professor does not give good notes, so I can't really follow his examples

8. Nov 20, 2012

### vish22

arildno has given you such big hints.
1.dxdy in polar becomes rdrd(theta)
2.(2+(x^2)+(y^2)) is your surface function,simply put x=rcos(theta) and y=rsin(theta) in the function of (x,y)-because theta is measured wrt. x-axis.
3.you have to think in terms of r and (theta).Your region (which is a quarter circle on the x-y plane-1st quadrant) can be covered completely if you vary r between 0 and 2 and (theta) between 0 and pi/2.
Try visualizing the situation.
PS:I may be wrong.

Last edited: Nov 20, 2012
9. Nov 20, 2012

### arildno

Remember that the area element is r*drd(theta).

The integral is radially symmetric, so it doesn't really matter if you integrate in the fourth quadrant or the first. (But the inequalities you gave for x and y denotes the first, not the fourth)

And, the limit on r is from 0 to 2

Last edited: Nov 20, 2012
10. Nov 20, 2012

### BenMcC

I ended up working it out and got an answer of 8pi+1/2. Not entirely confident in that answer

11. Nov 20, 2012