Question

Double Intergals in Polar Coordinates:

4. Use polar coordinates to find the volume of the solid that is bounded by the paraboloids
z = 3x^2 + 3y^2 and z = 4 ? x^2 ? y^2.

5. Evaluate by converting to polar coordinates ? -3 to3 * ? 0 to sqrt(9-x^2) (sin(x^2 +y^2) dydx

6. Evaluate by converting to polar coordinates: ? 0 to 1 * ? -sqrt(1-y^2) to 0 (x^2(y)) dxdy

Answer 1

Answer 2

(5)

..Almost that simple. Yes r^2 = x^2 + y^2 and dA = r dr d?.

What about the region of integration?
In this case, we have y = 0 to sqrt(9 - x^2) with x = -3 to 3.

Observe that this region is just the first quadrant of the disk x^2 + y^2 = 9
In polar coordinates, this is given by r in [0, 3] and ? in [0, ?/2].

Therefore, the double integral in polar coordinates equals
?(? = 0 to ?/2) ?(r = 0 to 3) sin(r^2) * r dr d?
= [?(? = 0 to ?/2) d?] * [?(r = 0 to 3) r sin(r^2) dr]
= ?/2 * [(-1/2) cos(r^2) {for r = 0 to 3}]
= (?/4) (1 - cos 9).

(4)

We work in polar coordinates. First we locate the bounds on (r, ?) in the xy-plane. The curve of
intersection of the two surfaces is cut out by the two equations z = 3 and x
2 + y^2= 1. Therefore
the x- and y- coordinates in this solid region must lie in the disk of radius one, i.e., where 0 ? r ? 1
and 0 ? ? ? 2?. To find the volume between the surfaces, we subtract the equation of the lower
surface,3x^2+3y^2=3r^2, from that of the higher one, integral(0 to 2 pie) integral (0 to 1)(4-4r^2)r dr dtheta =16pie/3...

Answer 3

(2)

..Almost that simple. Yes r^2 = x^2 + y^2 and dA = r dr d?.

What about the region of integration?
In this case, we have y = 0 to sqrt(9 - x^2) with x = -3 to 3.

Observe that this region is just the first quadrant of the disk x^2 + y^2 = 9
In polar coordinates, this is given by r in [0, 3] and ? in [0, ?/2].

Therefore, the double integral in polar coordinates equals
?(? = 0 to ?/2) ?(r = 0 to 3) sin(r^2) * r dr d?
= [?(? = 0 to ?/2) d?] * [?(r = 0 to 3) r sin(r^2) dr]
= ?/2 * [(-1/2) cos(r^2) {for r = 0 to 3}]
= (?/4) (1 - cos 9).

(1)

We work in polar coordinates. First we locate the bounds on (r, ?) in the xy-plane. The curve of
intersection of the two surfaces is cut out by the two equations z = 3 and x
2 + y
2 = 1. Therefore
the x- and y- coordinates in this solid region must lie in the disk of radius one, i.e., where 0 ? r ? 1
and 0 ? ? ? 2?. To find the volume between the surfaces, we subtract the equation of the lower
surface,3x^2+3y^2=3r^2, from that of the higher one, integral(0 to 2 pie) integral (0 to 1)(4-4r^2)r dr dtheta= 16 pie/3