Triple Integral Help: Solving Equations and Finding Volume with Closed Curve C

[Tarhead]

I have a group of problems that deals with the equations:

f(x,y)= x^2+y^2
g(x,y)=20-(x-4)^2-(y+2)^2.

I know that the surfaces z=f(x,y) and z=g(x,y) intersect in a closed curve, C, and the projection of C onto the xy-plane is a circle. However, I am having trouble finding its xy-equation, center, and radius. Additionally and more importantly, I am in the dark on setting up the double or triple integral for the volume of the region bounded by z=f(x,y) and z=g(x,y). Can anyone please help.

 

[Tide]

The two surfaces intersect along a circle centered at (x, y) = (2, -1) with a radius of \sqrt 5 so you might consider a coordinate transformation placing (2, -1) at the new origin.

 

[M98Ranger]

To find the xy-equation, center, and radius of the projection of C onto the xy-plane, we can use the fact that the projection of a circle onto the xy-plane is also a circle. This means that the xy-equation will also be a circle. We can write the equation as (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center of the circle and r is the radius.

To find the center, we can set both equations equal to each other and solve for x and y. This will give us the coordinates of the center. Once we have the center, we can find the radius by plugging in the coordinates of the center into either of the equations.

To set up the double or triple integral for the volume of the region bounded by z=f(x,y) and z=g(x,y), we first need to find the limits of integration. Since we know that the projection of C onto the xy-plane is a circle, we can use polar coordinates to set up the integral. The limits of integration for r would be from 0 to the radius of the circle, and for theta, it would be from 0 to 2π.

The integrand would be the difference between the two equations, g(x,y) and f(x,y). This will give us the height of the region at each point on the circle. Therefore, the integral would be ∫∫(g(x,y)-f(x,y))rdrdθ. This would give us the volume of the region bounded by z=f(x,y) and z=g(x,y).

I hope this helps. Let me know if you have any further questions. Good luck!