Asked by AIRA
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
1) solve using double integration of rectangular coordinate.
2) solve using double integration of polar coordinate
3)solve using triple intergation
1) solve using double integration of rectangular coordinate.
2) solve using double integration of polar coordinate
3)solve using triple intergation
Answers
Answered by
MathMate
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).
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).
Answered by
AIRA
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
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
Answered by
MathMate
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 <i>outside</i> 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.
The cylinder has a radius of 2 and its axis is along the z-axis.
So the volume of the paraboloid <i>outside</i> 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.
Answered by
AIRA
ohhh i see...
so my answer just now is wrong??
so my answer just now is wrong??
Answered by
Steve
yes, sorry I only gave the volume of the paraboloid, and forgot to subtract the cylinder inside. Better go with MathMate's answer.
Answered by
AIRA
it's ok.. thank you..
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