What is the Volume of a Paraboloid Between Two Planes?

[indie452]

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

 

[Cyosis]

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.

 

[Mark44]

[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.

 

[indie452]

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)

 

[Cyosis]

I find this pretty hard to explain without drawing pictures, but here goes.

$f(x,y)=x^2+y^2$. This is a paraboloid as stated and we can take $x^2+y^2 \leq z\leq 1$. 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 $\sqrt{1-x^2}$, thus $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$. All that remains now is the range for x, which is easy, take y=0 then $-1 \leq x \leq 1$.

This gives the integral:

$$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^1 dz dy dx$$

There is a simpler way, but I have to go now I'll be back in a few hours.

 

[indie452]

ok then just to check if you use the planes between z=0 and z=4 is the volume 8pi?

 

[Cyosis]

Yep that is correct you can use the formula $\frac{1}{2} \pi r h$ 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 $y=\sqrt{z}$ around the z-axis. now cut the paraboloid in slices, whose shapes are circles with radii $\sqrt{z}$. The area of one slice is therefore $\pi z$ adding all the circles together between the interval z=0 and z=1 yields the volume of the paraboloid.

$$\int_0^1 \pi z dz=\frac{\pi}{2}$$

Lots of text, but a very easy calculation.