Help me setup and solve double integral in polar coord.

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SUMMARY

The discussion focuses on setting up a double integral in polar coordinates to find the volume between the sphere defined by the equation x²+y²+z²=3 and the upper sheet of the hyperboloid z²=x²+y²+1. The curve of intersection is established as a circle of radius 1, leading to bounds of θ from 0 to 2π and r from 0 to 1. The integral is set up as ∫∫ (3 - (x² + y² + 1)) dA, which converts to polar coordinates as ∫∫ (2 - 2r²) r dr dθ, ultimately yielding a volume of V = π.

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  • Understanding of double integrals in calculus
  • Familiarity with polar coordinates and their conversion
  • Knowledge of spherical and hyperboloid equations
  • Ability to perform integration with respect to polar coordinates
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  • Learn about the properties of hyperboloids and their equations
  • Practice converting Cartesian coordinates to polar coordinates in integrals
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Students and educators in calculus, particularly those focusing on multivariable calculus, as well as anyone interested in understanding the application of polar coordinates in solving integrals.

RJLiberator
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Homework Statement


The region between sphere x^2+y^2+z^2=3 and the upper sheet of the hyperboloid z^2=x^2+y2+1.

Homework Equations

The Attempt at a Solution



Curve of intersection: We set the two equations equal to each other and find x^2+y^2=1, a circle of radius 1 is the curve of intersection.
I then drew the graph of the two elements here.

Now, I need to set up the actual double integral.
I know my bounds: theta from 0 to 2pi and r from 0 to 1.

I know that using polar coordinates I have to use r*dr*d(theta).

My problem is, I don't know how to set up the rest of the integral. I think it has something to do with f(r,theta)-g(r,theta) but I've been trying various things and nothing seems to be clear/working out.

I think I should do (3-x^2-y^2) - (x^2+y^2+1) as that seems to be the correct order, but how do I switch this to polar coordinates :/.
 
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After re-looking at some conversion formula's, I may have solved it:

So we do (3-x^2-y^-2)-(x^2+y^2+1) which is upper-lower and we get 2-2x^2-2y^2.
We then convert to polar coordinates and simplify the sin^2(theta)+cos^2(theta) so that we get: 2-2r^2
We can now use this in the integral with the bounds that I posted above, we multiply both by r: 2r-2r^3 and then we simply integrate.
The answer should be V = pi.

:)
 

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