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
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).
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.
ohhh i see...
so my answer just now is wrong??
yes, sorry I only gave the volume of the paraboloid, and forgot to subtract the cylinder inside. Better go with MathMate's answer.
it's ok.. thank you..
To find the volume of the solid inside the paraboloid, outside the cylinder, and above the xy-plane, we can approach it using three different methods: double integration of rectangular coordinates, double integration of polar coordinates, and triple integration.
1) Double Integration of Rectangular Coordinates:
To solve using rectangular coordinates, we first need to determine the limits of integration for x and y.
The paraboloid equation is given as z = 9 - x^2 - y^2, and the cylinder equation is x^2 + y^2 = 4.
By setting z = 0, we can find the limits of integration for both x and y.
9 - x^2 - y^2 = 0
x^2 + y^2 = 9
This represents the circle centered at the origin with radius 3. So, we can take the limits of integration for x as -3 to 3 and for y as -√(9-x^2) to √(9-x^2).
The volume can then be calculated using the double integral of the function 1 with respect to x and y over these limits.
2) Double Integration of Polar Coordinates:
To solve using polar coordinates, we can express the equations in terms of polar coordinates: x = r*cosθ and y = r*sinθ.
For the cylinder equation x^2 + y^2 = 4, we can substitute the values of x and y.
(r*cosθ)^2 + (r*sinθ)^2 = 4
r^2(cos^2θ + sin^2θ) = 4
r^2 = 4
This represents a circle centered at the origin with radius 2. So, we can take the limits of integration for r as 2 to 3 and for θ as 0 to 2π.
Next, we need to express the paraboloid equation in terms of polar coordinates. Using the substitutions for x and y, we have:
z = 9 - x^2 - y^2
z = 9 - r^2(cos^2θ + sin^2θ)
z = 9 - r^2
The volume can then be calculated using the double integral of the function r with respect to r and θ over these limits.
3) Triple Integration:
To solve using triple integration, we can directly integrate the function 1 over the volume enclosed by the paraboloid, cylinder, and xy-plane.
By setting up the bounds of the triple integral, we will integrate with respect to x, y, and z.
The limits of integration for x and y are the same as mentioned in the previous methods, i.e., -3 to 3 for x and -√(9-x^2) to √(9-x^2) for y.
For z, the bounds range from the xy-plane (z = 0) to the top of the paraboloid (z = 9 - x^2 - y^2). So, the limits of integration for z are 0 to 9 - x^2 - y^2.
The volume can be calculated using the triple integral of the function 1 with respect to x, y, and z over these limits.
Note: The integration process for each method will involve different steps depending on the coordinate system being used. It is important to verify the final answer by performing the calculations accurately.
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