Volume of a Region bounded by two surfaces

[RadiationX]

Find the volume of the solid region R bounded above by the paraboloid

$$z=1-x^2-y^2$$ and below by the plane $$z=1-y$$

The solution to this problem is:

$$V=\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}} (1-x^2-y^2)dxdy -\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}}(1-y)dxdy$$

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 $$\sqrt{1-x^2}$$ (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?

 

[HallsofIvy]

One thing you did not do is determne the intersection of the paraboloid and the plane. You need to do that to determine your limits of integration just as you need to find where two 2-d graphs intersect to determine the area between them.

The parabolid is z= 1- x²-y² and the plane is z= 1-y. Substituting that gives 1- y= 1- x²- y² or x²+y²+ y= 0. Completing the square, x²+ (y- 1/2)²= 1/4. That is, projecting the solid into the xy-plane, the boundary is a circle with center at (0,1/2) and radius 1/2 (in other words, it is tangent to the x-axis).

Clearly, y ranges from 0 up to 1. For each y, x ranges (solve x²= y²- y for x) $$-\sqrt{y-y^2}$$ to $$\sqrt{y-y^2}$$.
Those are the limits of integration.

 

[RadiationX]

One thing you did not do is determne the intersection of the paraboloid and the plane. You need to do that to determine your limits of integration just as you need to find where two 2-d graphs intersect to determine the area between them.

The parabolid is z= 1- x²-y² and the plane is z= 1-y. Substituting that gives 1- y= 1- x²- y² or x²+y²+ y= 0. Completing the square, x²+ (y- 1/2)²= 1/4. That is, projecting the solid into the xy-plane, the boundary is a circle with center at (0,1/2) and radius 1/2 (in other words, it is tangent to the x-axis).

Clearly, y ranges from 0 up to 1. For each y, x ranges (solve x²= y²- y for x) $$-\sqrt{y-y^2}$$ to $$\sqrt{y-y^2}$$.
Those are the limits of integration.

You are correct even the picture in the solution is a circle tangent to the x-axis. But why the need for two double integrals?

 

[HallsofIvy]

I would have written the integral itself as
$$V=\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}} (1-x^2-y^2- (1-y))dxdy =\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}}(y-x^2-y^2)dxdy$$,
integrating the distance between the two surfaces. What the formula you have does is find the volume from the paraboloid to the xy-plane, then the volume from the x= 1-y plane to the xy-plane and subtract to find volume between those.

Of course, since $$\int fdxdy- \int gdxdy = \int (f-g)dxdy$$,
it's exactly the same thing.