Triple Integral, Volume of an Egg

In summary: So it makes sense to integrate it from 0 to 4, since that is when the value of r is the smallest.Can you explain why I integrate dr from 0 to 4? Why not 0 to 3?I don't know. Maybe because it makes the function easier to graph? I don't know. Maybe because it makes the function easier to graph?
  • #1
ktvphysics
14
0

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.
 
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  • #2
ktvphysics said:

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.
 
  • #3
Yes, I agree. What you said is basically my progress so far.
 
  • #4
I'd do this as a volume of rotation.
Are you instructed to do it via a triple integral?
 
  • #5
Simon Bridge said:
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.
 
  • #6
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.
 
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  • #7
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?
 
  • #8
Simon Bridge said:
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.
 
  • #9
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?
 
  • #10
TheGeometrist said:
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
 
  • #11
ktvphysics said:
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.
 
  • #12
TheGeometrist said:
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?
 
  • #13
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?
 
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  • #14
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.
 
  • #15
Simon Bridge said:
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
TheGeometrist said:
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?
 
  • #16
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.
 
  • #17
Simon Bridge said:
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.
 
  • #18
I got 12pi for my final answer.

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

Related to Triple Integral, Volume of an Egg

What is a triple integral and how is it used?

A triple integral is a mathematical tool used to calculate the volume of a three-dimensional object, such as an egg. It involves integrating a function over a three-dimensional region and is often used in physics, engineering, and other scientific fields to calculate the mass, center of mass, and other important properties of objects.

How do you calculate the volume of an egg using a triple integral?

To calculate the volume of an egg using a triple integral, you first need to determine the boundaries of the three-dimensional region that represents the egg. This can be done by finding the equations of the egg's surface and setting them as the boundaries of the integral. Then, you integrate the function representing the egg's volume over these boundaries to obtain the total volume.

Can a triple integral be used to calculate the volume of any three-dimensional object?

Yes, a triple integral can be used to calculate the volume of any three-dimensional object, as long as the boundaries of the integral can be determined. This technique is especially useful for irregularly shaped objects, such as eggs, where other methods may not be as accurate.

What are some real-world applications of using a triple integral to calculate the volume of an egg?

One real-world application of using a triple integral to calculate the volume of an egg is in the food industry. By accurately calculating the volume of an egg, companies can determine the amount of egg needed for a particular recipe or product. This can help them save money and reduce waste. Triple integrals are also used in engineering to calculate the volume of complex structures, such as airplane wings or car engines.

Are there any limitations to using a triple integral to calculate the volume of an egg?

While triple integrals are very useful for calculating the volume of three-dimensional objects, they do have some limitations. For example, they may not be suitable for very complex or irregularly shaped objects where the boundaries of the integral are difficult to determine. In addition, the accuracy of the volume calculation may depend on the accuracy of the equations used to represent the egg's surface.

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