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Finding a volume in space

  • Thread starter hyper
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  • #1
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Hello, I need to find the volume between the plane z=1 and z=10-x*x-4*y*y.

I have tried using polar coordinates:
first integrate from z=1 to z=10-x*x-4*y*y
The I integrate in the plane z=1, but here I need to integrate over an ellipse, how do I do this? When I wrote the expression to polar coordinates I got an expression for r that I couldnt integrate.

I have also tried integrating over the ellipse using cartesian coordinates, but I got stuck aswell, can someone pleas help?
 

Answers and Replies

  • #2
Cyosis
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Could you show the actual function you have to integrate over? I suspect it has the form [itex]\sqrt{9-4y^2}[/itex]? If so use the substitution [itex]u=\frac{3}{2} \sin \theta[/itex]. Still integrating like this isn't very pretty. A more elegant method and by far the easiest to integrate is to notice that this is a paraboloid with elliptical cross section. You can then slice the paraboloid parallel to the x-y plane in a lot of slices and then integrate from z=1 to 10. The only thing you have to do is find an expression for the semi major axis and the semi minor axis as a function of z and use that the area of an ellipse is given by pi a b, with a the semi major axis and b the semi minor axis.
 
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  • #3
HallsofIvy
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Could you show the actual function you have to integrate over? I suspect it has the form [itex]\sqrt{9-4y^2}[/itex]? If so use the substitution [itex]u=\frac{3}{2} \sin \theta[/itex]. Still integrating like this isn't very pretty. A more elegant method and by far the easiest to integrate is to notice that this is a paraboloid with elliptical cross section. You can then slice the paraboloid parallel to the x-y plane in a lot of slices and then integrate from z=1 to 10. The only thing you have to do is find an expression for the semi major axis and the semi minor axis as a function of z and use that the area of an ellipse is given by pi a b, with a the semi major axis and b the semi minor axis.
No, finding the volume bounded by z= 1 and [itex]z= 10- x^2- 4y^2[/itex] involves integrating the difference in z values, [itex](10- x^2- 4y^2)- 1= 9- x^2- y^2[/tex] over the circle [itex]x^2+ 4y^2= 9[/itex], where the two surfaces cross.

Now the limits of integration, in xy-coordinates will involve something like [itex]y= \frac{1}{2}\sqrt{9- x^2}[/itex].

Hyper, you might consider using parameters r and [itex]\theta[/itex] such that [itex]x= r cos(\theta)[/itex] and [itex]y= 4r sin(\theta)[/itex], not standard polar coordinates, with r going from 0 to 1 and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. Do you know how to find the "differential of area" in r and [itex]\theta[/itex] using the Jacobian?
 
  • #4
Cyosis
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Why wouldn't slices be applicable, it yields the same answer?
 

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