What is the Volume of a Paraboloid Between Two Planes?

In summary, the conversation revolves around finding the volume of a paraboloid between the planes z=0 and z=1. The solution involves rearranging the function and performing a triple integral with limits for z, y, and x. There are multiple approaches to solving this problem, including using the formula for the volume of a cone and taking advantage of symmetry. The final answer is pi/2.
  • #1
indie452
124
0

Homework Statement



evaluate volume of paraboloid z = x2 + y2 between the planes z=0 and z=1

The Attempt at a Solution



i figured we would need to rearrange so that F(x,y,z) = x2 + y2 - z

then do a triple integral dxdydz of the function F. the limits for the first integral dz would be z=1 and z=0. and i don't know what the other limits would be (y1,y2 and x1, x2?)but this first integral gave an answer of -1/2. this would mean that the volume would end up being zero which i don't think is right.

then i thought that maybe i should say the function is z(x,y) = x2 + y2 and integrating dxdy.

the dy limits would then be
[when z=1] y2 = 1 - x2
y = sqrt[1-x2]

[when z=0] y2 = -x2
y = sqrt[-x2] = xi i=complex numberthe dx limits would then be
[when z=1] x2 = 1 - y2
x = sqrt[1-y2]

[when z=0] x2 = -y2
x = sqrt[-y2] = yi i=complex numberbut this seems like a dead end
any suggestions would be helpful

ok just tried another thing: the lower limits for dx and dz being zero taking out the complex numbers.

this gives: after dy=> 2y dx = 2sqrt(1-x2)dx = x*sqrt(1-x2) + sin-1x

i then put in the limits 1 an 0 and got pi/2

still don't know if this is right though
 
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  • #2
The volume is pi/2, but I can't say I understand how you arrived at that answer. To check if your expression is correct you could take the planes z=0 and z=2, which should give you a volume 2pi.
 
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  • #3
[when z=0] y2 = -x2
y = sqrt[-x2] = xi i=complex number
Not true. When z = 0, x2 + y2 = 0, which means that both x and y are 0.
 
  • #4
Cyosis said:
The volume is pi/2, but I can't say I understand how you arrived at that answer. To check if your expression is correct you could take the planes z=0 and z=2, which should give you a volume 2pi.

~ ok well i tried it and got pi so I am not sure what i have done wrong...

~ i have basically done a double integral dxdy (ie dy was done first then dx of that)...

~ this gave 2y which i then needed to sub in the limits.
when z=0, y=0 so this is the lower limit
when z=2, y= sqrt(2-x2)

~ this gave me 2sqrt(2-x2), this integrates to give:
2*[sin-1(x/sqrt(2)) + (x/2)(sqrt(2-x2))]

~ the limits for this are
when z=0, x=0 so this is the lower limit
when z=2, y= sqrt(2)
 
  • #5
I find this pretty hard to explain without drawing pictures, but here goes.

[itex]f(x,y)=x^2+y^2[/itex]. This is a paraboloid as stated and we can take [itex]x^2+y^2 \leq z\leq 1[/itex]. If we slice this paraboloid in small slices parallel to the x-y plane we get a lot of circles. These can be described as [itex]\sqrt{1-x^2}[/itex], thus [itex]-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}[/itex]. All that remains now is the range for x, which is easy, take y=0 then [itex]-1 \leq x \leq 1[/itex].

This gives the integral:

[tex]\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^1 dz dy dx[/tex]

There is a simpler way, but I have to go now I'll be back in a few hours.
 
  • #6
ok then just to check if you use the planes between z=0 and z=4 is the volume 8pi?
 
  • #7
Yep that is correct you can use the formula [itex]\frac{1}{2} \pi r h[/itex] with r the radius of the greatest circle (top or bottom of the parabola depending on its orientation) and h its height.

While the method in my previous post will work for general cases sometimes looking at symmetry and clever thinking saves you a lot of computing time. For example in this case where the problem is completely rotational symmetric around the z-axis. You can obtain this paraboloid by simply rotating [itex]y=\sqrt{z}[/itex] around the z-axis. now cut the paraboloid in slices, whose shapes are circles with radii [itex]\sqrt{z}[/itex]. The area of one slice is therefore [itex]\pi z[/itex] adding all the circles together between the interval z=0 and z=1 yields the volume of the paraboloid.

[tex]\int_0^1 \pi z dz=\frac{\pi}{2}[/tex]

Lots of text, but a very easy calculation.
 

1. What is a paraboloid?

A paraboloid is a three-dimensional geometric shape that resembles a parabola. It is a surface formed by rotating a parabola around its axis.

2. How do you calculate the volume of a paraboloid?

The formula for calculating the volume of a paraboloid is V = (π/2) * r^2 * h, where r is the radius of the base and h is the height.

3. What are the applications of paraboloids in science?

Paraboloids are commonly used in optics, such as in satellite dishes and telescopes, as they can focus and reflect light. They are also used in engineering for designing structures that can withstand pressure, such as in cooling towers and water tanks.

4. How does the volume of a paraboloid change with respect to its dimensions?

The volume of a paraboloid is directly proportional to the square of its radius and its height. This means that if the radius or height is doubled, the volume will increase by a factor of four. Similarly, if the radius or height is halved, the volume will decrease by a factor of four.

5. How is the volume of a paraboloid different from a sphere?

A paraboloid and a sphere have different shapes and therefore, different formulas for calculating their volumes. While a paraboloid has a flat base and a curved surface, a sphere has a perfectly curved surface. Additionally, the formula for calculating the volume of a sphere is V = (4/3) * π * r^3, where r is the radius.

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