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# polar coordinates

Use double integration in polar coordinates to find the volume of the solid that lies below the given surface and above the plane region R bounded by the given curve.

1. z=x^2+y^2; r=3

Evaluate the given integral by first converting to polar coordinates.

2. ∬_(0,x)^1,1▒〖x^2 dy dx〗

Solve by double integration in polar coordinates.

3. Find the volume of the solid bounded by the paraboloids z=12-2x^2-y^2 and
z=x^2+2y^2

Find the centroid of the plane region bounded by the given curves. Assume that the density is δ≡1 for each region.

4. x=-2, x=2, y=0, y=x^2+1

Find the mass and centroid of the plane lamina with the indicated shape and
density.

5. The region bounded by x=e, y=0, and y=ln⁡x for 1≦x≦e,with
δ(x,y)≡1

#### Solution Preview

Problems:

1. Use double integration in polar coordinates to find the volume of the solid that lies below the given surface and above the plane region R bounded by the given curve.

2. Evaluate the given integral by first converting to polar coordinates.

Solve by double integration in polar coordinates.

3. Find the volume of the solid bounded by the paraboloids z = 12 - 2x^2-y^2 and
z = x^2 + 2y^2

4. Find the centroid of the plane region bounded by the given curves. Assume that
the density is =1 for each region.
x = -2, x = 2, y = 0, y = x^2+1
5. Find the mass and centroid of the plane lamina with the indicated shape and
density. The region bounded by x = e, y = 0, and y = ln x for 1xe, with (x,y)=1

Solutions:

1. The general formula for calculation of volumes bounded by given surfaces is:
(1)
The plane region R where the curve r = ...

#### Solution Summary

This solution utilizes double integration in polar coordinates.

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