Calculating Volume Between Paraboloids and Cylinder

In summary, the conversation discusses how to find the volume of a solid between two paraboloid surfaces and inside a cylinder. The process involves determining which surface is on top and which is below, finding the intersection points, and using this information to set up the volume integral. It is also recommended to use a more convenient system of coordinates for integrating over a circular region. Additionally, it is important to check if any part of the intersection of the surfaces lies inside the cylinder in order to determine the correct bounds of integration.
  • #1
alexcc17
48
0
Find the exact volume of the solid between the paraboloids [itex]z=2x ^{2}+y ^{2}[/itex] and [itex]z=8-x ^{2}-2y ^{2}[/itex] and inside the cylinder [itex]x ^{2}+y ^{2}=1[/itex].

I really don't know how to set this up. Would it be something like ∫∫(2x^2+y^2)-(8-x^2-2y^2)dA + ∫∫(x^2+y^2-1)dA ?

If so, how would I find the bounds of integration?
 
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  • #2
Just like integrating areas between two curves in the plane, you need to determine additional information. First, which surface is on top and which surface is below? Rigorously, which function's values are greater in the region we are interested in? This is important for which volume is subtracted from which, as we want the volume of a solid to be positive. If you are not familiar with the equations of paraboloids, graph some cross sections to get a picture of the surface. If you have access to a 3d-graphing system, use that to verify your answer.
Next, where do the surfaces intersect within the region of interest, if anywhere? If they intersect within the region you are integrating over (the circle x^2 + y^2 = 1), you may need to break up the volume integral, as the volume changes sign if the surface that used to be below crosses the second surface and is now on top (just like when curves cross each other when doing area integrals).
Finally, use your information to determine the region in the xy-plane over which you will integrate the volume under the surfaces. If they do not intersect, you are simply integrating over the circular region x^2+y^2 = 1, which covers a cylindrical region of volume. Once you set up your integral correctly, you may want to change to a more convenient system of coordinates for integrating over a circular region, such as cylindrical coordinates.
 
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  • #3
What he said :)

If you already know how to find the volume under a surface z=f(x,y) inside a cylinder, the this problem is likely a matter of doing exactly that for each surface and finding the difference. The likely gotcha is if the surfaces themselves intersect inside the cylinder.

It is often easier to change coordinates to suit the symmetry.
 
  • #4
Thanks guys! z=8-x^2-2y^2 is on top

So would it be... ∫∫(8-x^2-2y^2)-(2x^2+y^2)dA + ∫∫(x^2+y^2-1)dA ?

I get how to solve the function for the volume, but typically I would find the bounds by setting the surfaces equal to each other to find intersection points and such, but setting them equal...
I get 9=4y^2 + 4x^2
 
  • #5
So the intersection is a circle at a constant z?! (you should always relate your equations to geometry.)
1. Check to see if any part of that intersection lies inside x^2+y^2=1 (the cylinder).

if none of it does - then the limits are entirely determined by the bounds of the cylinder instead of the intersection curve. (You realize you can do a volume integral between any limits you like - you are not restricted to the region between two surfaces?)
 
  • #6
Alright! So, I found that...
[itex]\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x ^{2}}}^{\sqrt{1-x ^{2}}}8-3x ^{2}-3y ^{2}dydx[/itex]
I found the square root part because [itex]y=\sqrt{1-x ^{2}}[/itex]

Is this correct, so all I have to do now is solve?

Thank you
 
  • #7
That looks like the right shape - you understand that I'm not going to actually do it unless someone pays me :)

Did you check to see if any of the intersection of the surfaces was inside the cylinder in question?
 

Related to Calculating Volume Between Paraboloids and Cylinder

1. What is the definition of volume in the context of double integrals?

In mathematics, volume refers to the measure of the amount of space occupied by a three-dimensional object. In the context of double integrals, volume is represented by a numerical value that is obtained by integrating a function over a two-dimensional region.

2. How is the concept of volume related to double integrals?

Double integrals are used to calculate the volume of a three-dimensional object by breaking it down into infinitesimally small rectangular slices. By integrating the area of these slices over a given region, the total volume can be obtained.

3. What is the formula for calculating volume using double integrals?

The formula for calculating volume using double integrals is given by V = ∬R f(x,y) dA, where R is the region of integration and f(x,y) is the function being integrated.

4. Can double integrals be used to calculate the volume of irregularly shaped objects?

Yes, double integrals can be used to calculate the volume of irregularly shaped objects by dividing the region of integration into smaller, simpler shapes and integrating over each of them separately.

5. What are the applications of double integrals in real-world problems?

Double integrals have many applications in different fields such as physics, engineering, and economics. They can be used to calculate the volume of complex objects, determine the mass and center of mass of an object, and even solve optimization problems.

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