Triple Integral, Volume of an Egg

[ktvphysics]

Homework Statement

I need to find the volume of an egg with a shape described by:

z = 1/2(x2 + y2) and z = 6 - x2 - y2

I am also given that the egg is 6cm in length.

Homework Equations

I roughly graphed the two surfaces. The first being paraboloid that opens up from the origin, and the second being a paraboloid that opens down from z =6. The region under these two surfaces is the "egg."

The Attempt at a Solution

I don't know where to begin to set this up. I know that the region I need is the region in between the two curves. I assume I integrate dz from 0 to 6. I don't know where to go from there.

 

[SteamKing]

Homework Statement

I need to find the volume of an egg with a shape described by:

z = 1/2(x2 + y2) and z = 6 - x2 - y2

I am also given that the egg is 6cm in length.

Homework Equations

I roughly graphed the two surfaces. The first being paraboloid that opens up from the origin, and the second being a paraboloid that opens down from z =6. The region under these two surfaces is the "egg."

The Attempt at a Solution

I don't know where to begin to set this up. I know that the region I need is the region in between the two curves. I assume I integrate dz from 0 to 6. I don't know where to go from there.

Sometimes, it's helpful to make a rough sketch of the two regions. The volume contained within the 'egg' is going to be under the upper surface, but above the lower surface, so this gives an indication on how to set up the z portion of the volume.

 

[ktvphysics]

Yes, I agree. What you said is basically my progress so far.

 

[Simon Bridge]

I'd do this as a volume of rotation.
Are you instructed to do it via a triple integral?

 

[ktvphysics]

I'd do this as a volume of rotation.
Are you instructed to do it via a triple integral?

Yes, I must solve with a triple integral.

 

[TheGeometrist]

It may help if you also try to sketch what slices parallel to the xy plane look like, to get a better idea for the parameters. In addition, perhaps you may consider a different coordinate system than Cartesian.

 

[ktvphysics]

I know that the intersection of the two surfaces forms a plane that is a circle of radius 4. (I set the z equations equal to each other and found x^2 + y^2 = 4). Is every cross sectional area a circle, or is that only in this instance? After thinking about it, I am retracting my statement in the original post about integrating dz from 0 to 6. I should integrate dz from z=1/2(x^2+y^2) to z=6-x^2-y^2, correct?

 

[ktvphysics]

I'd do this as a volume of rotation.
Are you instructed to do it via a triple integral?

I think I know how to solve it using a triple integral now. Solve it as a volume of rotation and we can compare answers, if you would like.

 

[TheGeometrist]

At any cross section parallel to the xy plane, you can consider z to be a constant (i.e. at z=3) with only x and y varying to make the shape. So If you consider z to be any constant, what do the equations z=6-x^2-y^2 and z=1/2(x^2+y^2) become, for any arbitrary constant value of z?

 

[ktvphysics]

At any cross section parallel to the egg, you can consider z to be a constant (i.e. at z=3) with only x and y varying to make the shape. If you consider z to be any constant, what do the equations z=6-x^2-y^2 and z=1/2(x^2+y^2) become, for any given value?

Circles. Got it. So is this how I set it up?

∫∫∫ rdzdrdΘ , integrating dz from z= 1/2(x^2+y^2) to z = 6 - x^2 - y^2, dr from 0 to 4 and dΘ from 0 to 2π

 

[TheGeometrist]

Circles. Got it. So is this how I set it up?

∫∫∫ rdzdrdΘ , integrating dz from z= 1/2(x^2+y^2) to z = 6 - x^2 - y^2, dr from 0 to 4 and dΘ from 0 to 2π

Close. The integration values for z are still in terms of x and y, while you are integrating in Cylindrical coordinates. It seems otherwise correct to me.

 

[ktvphysics]

Close. The integration values for z are still in terms of x and y, while you are integrating in Cylindrical coordinates. It seems otherwise correct to me.

Ah, okay. I would have ran into that issue when I tried to solve it I guess. So x^2 + y^2 = r^2, yes?

Can you explain why I integrate dr from 0 to 4? Why not 0 to 3? what does the intersection of the surfaces have to do with it?

 

[TheGeometrist]

Actually, sorry, I made a mistake as far as the integration of dr. Remember that r is of course the radius, and dr is being integrated from the minimum value of the surface to the maximum value. We can agree that it has a minimum of 0 (i.e. At z= 0, 6). In this case, the top function would keep expanding in radius in the negative z direction, and the bottom function would keep expanding in radius in the positive k direction. Therefore, the maximum value will have to be where the two intersect. Now what would the radius be at this maximum value?

 

[Simon Bridge]

The intersection of the surfaces sets the maximum width of the volume - you should be able to see that from your diagram.
Try taking a slice through the z-x plane. This means it determines some of the limits of the integration.

Note: You would expect r to be a function of z, since this is the radius of the circle you get when you slice perpendicular to the z-axis.

 

[ktvphysics]

The intersection of the surfaces sets the maximum width of the volume - you should be able to see that from your diagram.
Try taking a slice through the z-x plane. This means it determines some of the limits of the integration.


Note: You would expect r to be a function of z, since this is the radius of the circle you get when you slice perpendicular to the z-axis.

I'm not sure what you mean by that last bit... The part about the slice through the z-x plane

Actually, sorry, I made a mistake as far as the integration of dr. Remember that r is of course the radius, and dr is being integrated from the minimum value of the surface to the maximum value. We can agree that it has a minimum of 0 (i.e. At z= 0, 6). In this case, the top function would keep expanding in radius in the negative z direction, and the bottom function would keep expanding in radius in the positive k direction. Therefore, the maximum value will have to be where the two intersect. Now what would the radius be at this maximum value?

The maximum value of the radius is 2, right? x^2 + y^2 = 4. sqrt(4) = 2. That was my mistake, I forgot to take the square root of 4. So I would integrate dr from 0 to 2, correct?

 

[Simon Bridge]

Integrate what from 0-2?
Go carefully back to your setup.
You can tell if you got it correct by considering that you know the equation for the area of a circle.

 

[ktvphysics]

Integrate what from 0-2?
Go carefully back to your setup.
You can tell if you got it correct by considering that you know the equation for the area of a circle.

Integrate the radius from 0 to 2.

 

[ktvphysics]

I got 12pi for my final answer.

∫∫∫ r dzdrdΘ. 1/2r^2≤z≤6-r^2, 0≤r≤2, 0≤Θ≤2π = 12 pi