Volume of region R between paraboloid and xy-plane

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SUMMARY

The volume of the region R between the paraboloid defined by the equation z = 4 - x² - y² and the xy-plane is calculated using a double integral. The correct limits of integration are determined by the intersection of the paraboloid with the xy-plane, resulting in a circular region defined by x² + y² = 4. By converting to polar coordinates, the volume can be efficiently computed, yielding a final result of 8π.

PREREQUISITES
  • Understanding of double integrals
  • Familiarity with polar coordinates
  • Knowledge of paraboloid equations
  • Ability to determine limits of integration in calculus
NEXT STEPS
  • Study the application of double integrals in volume calculations
  • Learn about converting Cartesian coordinates to polar coordinates
  • Explore the properties of paraboloids and their intersections with planes
  • Practice solving integrals involving circular regions
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable integration, as well as educators teaching volume calculations using integrals.

stanford1463
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Homework Statement



So my question is: what is the volume of the region R between the paraboloid 4-x^2-y^2 and the xy-plane?

Homework Equations



I know how to solve it, it is a triple integral, but how do you find the limits of integration?

The Attempt at a Solution


Do I set x=0, or y=0 to find the limits of integration of the triple integral to find the volume ? or? I'm completely lost.
 
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stanford1463 said:

Homework Statement



So my question is: what is the volume of the region R between the paraboloid 4-x^2-y^2 and the xy-plane?

Homework Equations



I know how to solve it, it is a triple integral, but how do you find the limits of integration?

The Attempt at a Solution


Do I set x=0, or y=0 to find the limits of integration of the triple integral to find the volume ? or? I'm completely lost.

The paraboloid is defined by an equation, but I don't see one in your problem description.

A double integral will do the trick. For the limits of integration think about how this paraboloid intersects the x-y plane. The trace in the x-y plane is a circle, right?
 
Mark44 said:
The paraboloid is defined by an equation, but I don't see one in your problem description.

A double integral will do the trick. For the limits of integration think about how this paraboloid intersects the x-y plane. The trace in the x-y plane is a circle, right?

Umm, isn't the equation just z=4-x^2-y^2 ?
Ok, thanks, I'll try a double integral, so for the limits, I would just set z=0, right? and solve for x and y?
 
Ok, for my limits, i got x is between 0 and 2, and y is between 0 and 2...so i did a double integral and got 16/3. Is that right?
 
stanford1463 said:
Ok, for my limits, i got x is between 0 and 2, and y is between 0 and 2...so i did a double integral and got 16/3. Is that right?
No.
If the region over which you are integrating were a square, with 0 <= x <=2 and 0 <= y <= 2, those would be the limits, but it isn't a square. The region R is a circle whose equation is x^2 + y^2 = 4.

If you divide this region into small squares of area dx * dy, how far up and down (i.e., in the y direction) does a column of squares extend? How far left and right (i.e., in the x direction) do the columns of squares extend? If you can answer these questions correctly, you'll have your limits of integration.
 
Would it be plus/minus sqrt(4-x^2) ?
 
stanford1463 said:
Would it be plus/minus sqrt(4-x^2) ?
Close. A given "vertical" stack ranges from y = -sqrt(4 - x^2) to y = +sqrt(4 - x^2). For the second question, the "vertical" stacks range from x = -2 to x = +2.

You might notice that I answered with equations, something I'm trying to make you mindful of.

So now you have your limits of integration. You can make things slightly easier by exploiting the symmetry of your integrand. Because it involves only even powers of x and y, you can reduce your region to just the portion that's in the first quadrant in the x-y plane, introducing a multiplying factor of 4 to correct.

Things get interesting now because you'll need to integrate some terms that involve square roots and such. These are doable, but tedious. You can save yourself some trouble by converting to polar coordinates, with x = r cos(theta), y = r sin(theta), z = z. The limits of integration are especially simple because the region over which you are integrating is a circular disk. Just remember that dx dy in rectangular coordinates becomes r dr d(theta) in polar coordinates.
 
Alright, I used polar coordinates, i think, I got 8pi.
 

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