# Math Help - Volume integral

1. ## Volume integral

Find the volume of the region bounded by the cone $x^2 + y^2 = z^2~ (z \ge 0)$ and the paraboloid $z =2x^2 + 2y^2$.

------

I'm having a hard time even visualising the integration region. The following is my attempt to plot the paraboloid (in blue) and the cone (in brown) from x=-1..1 and y=-1..1.

Would I do this in rect-coords or cylindrical or what?

2. Originally Posted by scorpion007
Find the volume of the region bounded by the cone $x^2 + y^2 = z^2~ (z \ge 0)$ and the paraboloid $z =2x^2 + 2y^2$.

------

I'm having a hard time even visualising the integration region. The following is my attempt to plot the paraboloid (in blue) and the cone (in brown) from x=-1..1 and y=-1..1.

Would I do this in rect-coords or cylindrical or what?
two surfaces intersect at $z=0, \frac{1}{2}.$ for $0 \leq z \leq \frac{1}{2}$ the cone is the top surface. so the volume is: $V= \int \int_R [\sqrt{x^2+y^2} - 2(x^2+y^2)] dA,$ where $R=\{(x,y): \ x^2+y^2 \leq \frac{1}{4} \}.$

using polar coordinates we get: $V=\int_0^{2 \pi} \int_0^{\frac{1}{2}}(r^2-2r^3) \ dr d \theta = \frac{\pi}{48}.$

3. Thanks, but I have some questions:

How did you find their intersection? Did you solve the quartic polynomial:

$x^2+y^2 = 4x^4+8x^2y^2+4y^4$ ?

Don't we need a triple integral here? Actually, I guess not since the integrand would be a constant "1" for volume. So you somehow used a double over R...

How did you determine that the integrand should be the difference between the surface functions, I.e. $z_1 - z_2$?

4. Originally Posted by scorpion007
Thanks, but I have some questions:

How did you find their intersection? Did you solve the quartic polynomial:

$x^2+y^2 = 4x^4+8x^2y^2+4y^4$ ?

Don't we need a triple integral here? Actually, I guess not since the integrand would be a constant "1" for volume. So you somehow used a double over R...

How did you determine that the integrand should be the difference between the surface functions, I.e. $z_1 - z_2$?
to find the intersection: $z^2=x^2+y^2=\frac{z}{2}.$ thus $2z^2=z$ and hence $z=0, \frac{1}{2}.$ to answer your second question, recall that the volume of the region $D$ bounded by the two surfaces:

$z=f(x,y), \ z=g(x,y),$ with $f(x,y) \leq g(x,y), \ \forall (x,y) \in R,$ is $\int \int_R (g(x,y) - f(x,y)) dA,$ where $R$ is the image of $D$ on the xy plane.

5. Ah yes! Thank you very much!