Double Integrals in Polar Coordinates

[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.

 

[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.

 

[SteamKing]

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.

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.

 

[geezer73]

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

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

 

[Incand]

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

sorry, edited my post quickly after. meant volume!

 

[geezer73]

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.

Can this be graphed in 2 dimensions?

 

[geezer73]

sorry, edited my post quickly after. meant volume!

So this involves the z axis?

 

[SteamKing]

Can this be graphed in 2 dimensions?

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.

 

[Incand]

Can this be graphed in 2 dimensions?

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.

 

[geezer73]

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.

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?

 

[geezer73]

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.

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 :-)

 

[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 there's less area where y is big then where it is small so we get a bit less.

 

[SteamKing]

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.