Double Integrals in Polar Coordinates

1. Mar 16, 2015

geezer73

I'm in the middle of the Great Courses Multivariable Calculus course. A double integral example involves a quarter circle, in the first quadrant, of radius 2. In Cartesian coordinates, the integrand is y dx dy and the outer integral goes from 0 to 2 and the inner from 0 to sqrt(4-y^2). In polar, the integrand is (r sin theta) r dr dtheta, with the outer going from 0 to pi/2 and inner from 0 to 2. The final result, in both forms, is 8/3.

My thinking was that this would describe the area of the quarter circle, but elementary geometry tells me the area should be pi. I'm trying to visualize what this value of 8/3 represents. Maybe this sounds stupid, but I could use some help.

2. Mar 16, 2015

Incand

When you calculate a integral like that you ain't calculating the area. you are calculating the volume beneath $f(x,y) = y$ in the quarter circle.
$\iint 1dxdy$ would be the area.

3. Mar 16, 2015

SteamKing

Staff Emeritus
If you were calculating just the area of a quarter circle, the double integral would just be:

A = ∫∫ dx dy, since dA = dx dy and ∫∫ dA ≡ A

But, the integral which is actually being evaluated here is

I = ∫∫ y dx dy = ∫∫ y dA

In effect, I is the first moment of the area of the quarter circle about the x-axis.

4. Mar 16, 2015

geezer73

Thanks for the quick reply. I'm still unable to visualize what the "area under y" looks like. What would it's graph be?

5. Mar 16, 2015

Incand

sorry, edited my post quickly after. meant volume!

6. Mar 16, 2015

geezer73

Can this be graphed in 2 dimensions?

7. Mar 16, 2015

geezer73

So this involves the z axis?

8. Mar 16, 2015

SteamKing

Staff Emeritus
Not really. Since f(x,y) = y for this integral, you would need to add a third dimension. The base of the figure is the quarter circle of radius 2 centered at the origin, with the heights of the surface above the base being z = y.

I think it's better to interpret this double integral as representing the first moment of area of the quarter circle about the x-axis. You don't give yourself a headache trying to visualize a 3-dimensional surface. There are more complicated integrals than this for which there are no easy visualizations.

9. Mar 16, 2015

Incand

Not really no, or at least i don't think so. But the surface f(x,y) = y would just be a plane. and you simply sum up the volume beneath it in the quarter circle.
Think of it as a circle in the plane with $x^2+y^2=2$ and a plane above it $z-y=0 , x \in R$.

When doing multivariable calculus I find you really need to learn to draw 3d graphs, it really helps (I just finished the course myself only a week ago). If you got a lecturer, watch how he/she draws it on the board and try to copy it.

10. Mar 16, 2015

geezer73

Thanks, but I'm still trying to visualize. It sounds like, if this value is a volume, its average height on the z axis would be about .85. Is this in the right ballpark?

11. Mar 16, 2015

geezer73

Thanks much. I have some 3D graphing software, so I'll spend some more time with it. It's just good to know that the integral was not as simple as a 2D area. I was losing some marbles trying to reconcile 8/3 with pi :-)

12. Mar 16, 2015

Incand

Yes that would be about right. 8/(3pi) should be the average height. you could think of it that y goes up to 2. and the average would be about 1 but since its a circle and not a square theres less area where y is big then where it is small so we get a bit less.

13. Mar 16, 2015

SteamKing

Staff Emeritus
For a quarter circle, the centroid is located at x-bar = y-bar = 4 R / (3 π), so the first moment of area M = A * y-bar = [π R2/ 4] * [4 R / (3 π)]

M = R3 / 3

When R = 2, then M = 23 / 3 = 8/3

The π's cancel out, which is why π doesn't show up in the evaluation of this double integral.