Finding Volume Under Cone Above Disk

[Shaybay92]

Homework Statement

Use polar coordinates to find the volume of the given solid:
Under the cone z = Sqrt[x^2 + y^2]
Above the disk x^2 + y^2 <= 4

2. The attempt at a solution
I tried using formatting but I couldn't get it right so I'll explain...I changed variables by making the upper and lower limit of the inner integral [-2,2], with the outer integral [0,2pi]. The inner integral became integral of r^2 because Sqrt[ x^2 + y^2 ] is r then multiply by the r in r dr d(theta)... So I got

[r^3/3] from [-2,2] which gave me 16/3. I then integrated with respect to theta from 0 to 2pi (is this correct? Is the disk around the origin??) and that gave me 32pi/3 but the answer was the original 16pi/3. This is why I think it should be integrated from 0 to pi but I can't see why because the disk lies in all 4 quadrants.

 

[LCKurtz]

Homework Statement

Use polar coordinates to find the volume of the given solid:
Under the cone z = Sqrt[x^2 + y^2]
Above the disk x^2 + y^2 <= 4

2. The attempt at a solution
I tried using formatting but I couldn't get it right so I'll explain...I changed variables by making the upper and lower limit of the inner integral [-2,2], with the outer integral [0,2pi]. The inner integral became integral of r^2 because Sqrt[ x^2 + y^2 ] is r then multiply by the r in r dr d(theta)... So I got

[r^3/3] from [-2,2] which gave me 16/3. I then integrated with respect to theta from 0 to 2pi (is this correct? Is the disk around the origin??) and that gave me 32pi/3 but the answer was the original 16pi/3. This is why I think it should be integrated from 0 to pi but I can't see why because the disk lies in all 4 quadrants.

To cover a disk of radius 2 in polar coordinates, r does not go from -2 to 2. By convention in polar coordinates you usually take r nonnegative. Try r from 0 to 2 and theta from 0 to 2pi.

 

[Shaybay92]

Ah I see, looks like I may need to do work on polar coordinates :/