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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Volume of a Region bounded by two surfaces

**Physics Forums | Science Articles, Homework Help, Discussion**