# Math Help - Finding the volume of the solid formed by paraboloids.

1. ## Finding the volume of the solid formed by paraboloids.

Hello,

Having some problems on how to graph and how to set up the integral to find the volume in this question.

It asks sketch the solid bounded by the parabolic cylinders z + x^2 = 1,
x - y^2 = -1, x + y^2 = 1, and the plane z = 0.
It also aks to find the volume of this solid.

I know how to manipulate the equations:
z = 1 - x^2, which is the function we will integrate.

z = 0 = 1 - x^2, therefore x = +1 and -1, are these the limits when integrating with respect to dx?

Also, y = (1-x)^1/2 and y = (1+x0^1/2, are these the limits for integrating with respect to dy?

Thank you for any help.

2. It is difficult to explain how to do this without a 3-d plot.
But you end up if you closed solid. You then divide this solid into to parts. One composed of $x-y^2=1,x-y^2=-1$ and the upper surface of, $z=1-x^2$. Then, you prodced to find,
$\int \int_S \int dV$, where $S$ is solid formed. Understand? I realize it is difficult to be clear on this example.
---
Below is the curve of the region.
Divide it into 2 parts (the green and red)
Thus, the volume is (respectively),
$\int_{A_1} \int \int_0^{1-x^2} dz\, dA+\int_{A_2}\int \int_0^{1-x^2} dz\, dA$
Recognize that these are type II regions of integration,
$\int_{-1}^1 \int_0^{1-y^2} \int_0^{1-x^2}dz\, dy\, dx$+ $\int_{-1}^1 \int_0^{-1+y^2} \int_0^{1-x^2} dz\, dy\, dx$

3. Perfect Hacker,

The method of solution you gave me for finding the volume of the solid formed by the three paraboloids is a bit advanced for were I am at. We do integration in 3D next. With that said, I took a different approach at solving the problem:

I still used the graph that you posted.

In one way you can add the volume of the left and the right as follows:

for dx limits from -1 to 0, and for dy limits y = -(1+x)^1/2 to y = (1+x)^1/2 for the equation of (z = 1-x^2). This is the left side.

Then you add the right side with dx limits from 0 to 1, and dy limits from
y = -(1-x)^1/2 to y = (1-x)^1/2 for z = 1-x^2 being the equation we integrate.

This method gets me to an integrals 2[(1-x^2)*(1+x)^1/2] with respect to x from -1 to 0, and the addition of integral 2[(1-x^2)*(1-x)^1/2] with respect to x from 0 to 1.

I need to know how to expand these equations before I can integrate. I looked up the integral in a calculator and it gave me the value, but I need to know how they expanded the equations first. This is were I am at.

The other method was to integrate z = 1-x^2 with dy limits -1 to 1, and dx limits from x = 1-y^2 to x = y^2 -1. After all subsitutions, I ended up with an integral of 1/3(-2y^6 + 6t^4 -4) with respect to y from -1 to 1. I ended up with a negative volume of -224/105. I have a feeling this is right but cannot confirm it as I do not know if the solid goes below the xy plane.

Thanks again.

4. If you are not allowed to use triple integral you can do it in doubles.

Again, divide you solid into two parts the green and red.
Under the surface, $z=1-x^2$,
Thus,
$
\int_{A_1} \int 1-x^2 dA+\int_{A_2}\int 1-x^2 dA
$

Now, this is a type II region thus,
$
\int_{-1}^1 \int_0^{1-y^2} 1-x^2 dx\, dy$
+ $\int_{-1}^1 \int_{-1+y^2}^0 1-x^2 dx\, dy
$

Note: I made a mistake before. Because first I wrote,
$\int_0^{-1+y^2}$ but I should have written,
$\int_{-1+y^2}^0$ because the left side in this region is a curve first and then the vertical line $x=0$. (The answer I get is approximately 2.06)

5. Thank you again Perfect Hacker.

The double integral methos you proposed works. I got the same answer as you did, and you can also get the same answer by the method I proposed were I asked you how to expand the expression that I looked up the integral for.

Thank you very much for all your help. Very kind.

jcarlos