find the volume of solid inside the paraboloid z=9-x^2-y^2, outside the cylinder x^2+y^2=4 and above the xy-plane
The cylinder intersects the paraboloid at z=5.
The volume inside the paraboloid is
v = ∫[0,2π] ∫[0,2] ∫[0,9-r^2] r dz dr dθ
= ∫[0,2π] ∫[0,2] r(9-r^2) dr dθ
= ∫[0,2π] 14 dθ
= 28π
To find the volume of the solid inside the paraboloid, we need to set up a double integral over the region that satisfies the given conditions.
First, let's find the limits of integration for x and y. The cylinder equation tells us that x^2 + y^2 = 4, which represents a circle with a radius of 2. To visualize this, imagine a circular region on the xy-plane centered at the origin with a radius of 2.
Now, we want to find the volume above the xy-plane, so the z-values will range from z = 0 (the xy-plane) to the z-value given by the paraboloid equation z = 9 - x^2 - y^2.
To set up the double integral, we need to express the given region as a combination of the limits of integration for x and y. We can describe this region as follows:
For x, the limits will be from -2 to 2 (since the circle has a radius of 2).
For y, the limits will be from the lower semicircle to the upper semicircle of the circle.
So, the limits for y will be from -√(4-x^2) to √(4-x^2).
Now, let's set up the double integral:
∫∫R (9 - x^2 - y^2) dy dx,
Where R represents the region we described earlier.
Evaluating this double integral will give us the volume of the solid inside the paraboloid, outside the cylinder, and above the xy-plane.