Finding the Volume of a Solid Below a Plane and Above a Paraboloid

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Homework Help Overview

The problem involves finding the volume of a solid that is situated below the plane defined by the equation z=2x and above the paraboloid given by z=x^2 + y^2. The original poster seeks assistance in setting up the problem for evaluation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the importance of visualizing the problem through contour lines and question the setup of the equations involved. There is a focus on determining the domain for integration and the conditions under which the two surfaces intersect.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Guidance has been offered regarding the setup of integrals and the use of polar coordinates, which may help in progressing towards a solution.

Contextual Notes

There is mention of specific conditions for integration and the need to identify the intersection points of the surfaces involved. The original poster expresses uncertainty about the initial setup and the implications of setting z=0.

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Homework Statement


The volume of the solid below the plane: z=2x and above the paraboloid z=x^2 + y^2.

I need help setting this one up, I can handle the evaluating.


The Attempt at a Solution



I just don't know.
 
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Drawing a picture always helps. Try looking at a few contour lines.
 
That's where i am having a problem... I believe you are suppose to set Z to 0. But then X^2 + y^2 = 0... So x and y =0?
 
That is one of the contour lines, yes. It's also one of the endpoints on your integral. Where is the domain that you integrate over? Hint: It's where z1=z2, where z1=2x and z2=x^2+y^2.
 
I'm assuming after you set z1=z2. You solve for each? Setting the other one to zero. So x=2 and y = 0?

Is it:

2 y...2 y
S S 2x dxdy - S S x^2 + y^2 dxdy
0 0...0 0
?
 
Well, consider that you're integrating over the circle (x-1)^2+y^2 = 1. Considering that, I might go into polar coordinates...
 
thanks very much. I think i got it from here.
 
No problem. Glad I could help.
 

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