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

Question

1) solve using double integration of rectangular coordinate.
2) solve using double integration of polar coordinate
3)solve using triple intergation

Answer 1

First find out where z(x,y) intersects the x-y plane, which turns out to be a circle given by
x²+y²=9
So you would need evaluate the volume of the paraboloid above the x-y plane between the circles
x²+y²=2² (cylinder, radius = 2)
and
x²+y²=3² end of volume above x-y plane, r=3.

In polar coordinates, it would be
∫[0,2π]∫[2,3] z(x,y)rdrdθ
z(x,y) can be converted to r,θ by the substitution
x=rcosθ
y=rsinθ
so
z(x,y)=9-x²-y²
=9-r²(cos²θ+sin²θ)
and the integral becomes
Volume
=∫[0,2π]∫[2,3] (9-r²)r dr dθ
=∫[0,2π]∫[2,3]
(9r-r³) dr dθ
=2π[9r²/2-r^4/4] [2,3]
=2π(25/4)
=25π/2

In rectangular coordinates, you could integrate over a semi-annulus from -3 to +3.

Triple integration would be similar to (1) and (2), where z goes from 0 to z(x,y) or z(r,θ).

Answer 2

Why not this kind of answer?
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

Answer 3

Unless I am mistaken, the paraboloid looks like a mountain, tapering off to zero along a circle of radius 3.

The cylinder has a radius of 2 and its axis is along the z-axis.

So the volume of the paraboloid outside the cylinder is like you drilled a vertical hole of radius 2 through the mountain (of radius 3).

I hope you can visualize the situation.

Answer 4

ohhh i see...
so my answer just now is wrong??

Answer 5

yes, sorry I only gave the volume of the paraboloid, and forgot to subtract the cylinder inside. Better go with MathMate's answer.

Answer 6

it's ok.. thank you..