Calculating Volume of a Region: Integration in Polar Coordinates

[Calpalned]

Homework Statement

2. Homework Equations
n/a

The Attempt at a Solution

[/B]1) Whenever I need to find the volume between two surfaces, the integrand is simply the difference (subtraction) of the two equations? In the solution guide above, it is clear that they subtracted the two equations for z.

2) After transforming the double integral into polar coordinates, how did the solutions guide figure out the limits of integration? The object being integrated is a paraboloid limited by z = 4. Why then do the limits go from 0 to 2pi? Where do the limits for R (the inner integral) come from?

 

[HallsofIvy]

1) Whenever I need to find the volume between two surfaces, the integrand is simply the difference (subtraction) of the two equations? In the solution guide above, it is clear that they subtracted the two equations for z.

You shouldn't have to ask! You can take a small "delta x- delta y" rectangle in the xy-plane and then the height of the rectangular solid is the z distance between the bottom and the top- that is, the difference between "the two equations". The volume is z delta x delta y which, in the limit becomes the integral of the z difference dx dy.

2) After transforming the double integral into polar coordinates, how did the solutions guide figure out the limits of integration? The object being integrated is a paraboloid limited by z = 4. Why then do the limits go from 0 to 2pi? Where do the limits for R (the inner integral) come from?

In your uv- coordinates the two bounding surfaces are z= 4 and z= u^2+ v^2. They intersect at u^2+ v^2= 4. You should be able to recognize that as a circle in the uv- plane with center at (0, 0) and radius 2. To cover that circle, take r from 0 to 2 and $\theta$ from 0 to $2\pi$.