Find the volume of the solid region R bounded above by the paraboloid(adsbygoogle = window.adsbygoogle || []).push({});

[tex]z=1-x^2-y^2[/tex] and below by the plane [tex]z=1-y[/tex]

The solution to this problem is:

[tex]V=\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}} (1-x^2-y^2)dxdy -\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}}(1-y)dxdy[/tex]

I thought that i understood how to solve these types of problems but I don't understand this solution. This is the procedure that I would use to solve this problem, please correct my mistakes.

1. set z=o in the paraboloid. this reduces the problem from 3-d to 2-d

2. graph the curve of the paraboloid in 2-space (it's a circle)

3. set z=0 in the plane. this reduces the plane to 2-space ( it's a line)

4 what i have now is something like a half circle or hemisphere

5. find the limits of integration in the x and y directions

Here is where my solution fails. If y runs from the lower limit of 1 to the upper limit of [tex]\sqrt{1-x^2}[/tex] (top of paraboloid), how do i find the limits of x?

As another note I thought that I could use one double integral to solve this. What type of volume problem requires two double integrals as opposed to one double integral?

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# Volume of a Region bounded by two surfaces

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