# Math Help - Triple Integral

1. ## Triple Integral

Let A be the region that in space bounded by the balls:
$x^2 +y^2 + z^2 =1$ , $x^2 +y^2 +z^2 =4$ , above the plane $z=0$ and inside the cone $z^2 = x^2 +y^2$.
A. Write the integral $\int \int \int_{A} f(x,y,z) dxdydz$ in the form:
$\int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z) dz) dxdy$ when :
$A=( (x,y,z) | (x,y) \in E, g^1(x,y) \le z \le g^2(x,y) )$ ...
B. Find the volume of A (not necessarily using part A).
Hope you'll be able to help me in this... I think the main problem is that I can't figure out how A looks like... There is also a hint that one of the functions g1 or g2 should be defined at a split region... I can't figure out how this cone looks like and how I can describe A as equations ...

2. Originally Posted by WannaBe
Let A be the region that in space bounded by the balls:
$x^2 +y^2 + z^2 =1$ , $x^2 +y^2 +z^2 =4$ , above the plane $z=0$ and inside the cone $z^2 = x^2 +y^2$.
A. Write the integral $\int \int \int_{A} f(x,y,z) dxdydz$ in the form:
$\int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z) dz) dxdy$ when :
$A=( (x,y,z) | (x,y) \in E, g^1(x,y) \le z \le g^2(x,y) )$ ...
B. Find the volume of A (not necessarily using part A).
Hope you'll be able to help me in this... I think the main problem is that I can't figure out how A looks like... There is also a hint that one of the functions g1 or g2 should be defined at a split region... I can't figure out how this cone looks like and how I can describe A as equations ...

The cone intersects the inner sphere, $x^2+ y^2+ z^2= 1$ where $x^2+ y^2+ (x^2+ y^2)= 2(x^2+ y^2)= 1$, the circle of radius $\sqrt{2}/2$, in the plane z= 1/2, with center at (0, 0, 1/2). It intersects the outer sphere, $x^2+ y^2+ z^2= 4$ where $x^2+ y^2+ (x^2+ y^2)= 2(x^2+ y^2)= 4$, the circle of radius $\sqrt{2}$, in the plane z= 2, with center at (0, 0, 2).

The volume inside the cylinder, $x^2+ y^2= 1/2$, is easy:
$\int_{x= -\sqrt{2}/2}^{\sqrt{2}/2}\int_{y= -\sqrt{1/2- x^2}}^{\sqrt{1/2- x^2}}\int_{z=\sqrt{1- x^2- y^2}}^{\sqrt{1- x^2- y^2}} dzdydx$.

But that misses the volume, above the cone, that lies in the annulus between the two circles, $x^2+ y^2= 1$ and $x^2+ y^2+ z^2= 4$. To do that you will need to separate it into four integrals:

The first with x from -2 to $-\sqrt{2}/2$, y from $-\sqrt{4- x^2}$ to $\sqrt{4- x^2}$, z from $\sqrt{x^2+ y^2}$ to $\sqrt{4- x^2- y^2}$

The second with x from $-\sqrt{2}/2$ to $\sqrt{2}/2$, y from $\sqrt{1- x^2}$ to $\sqrt{4- x^2}$, z from $\sqrt{x^2+ y^2}$ to $\sqrt{4- x^2- y^2}$

The third with x from $-\sqrt{2}/2$ to $\sqrt{2}/2$, y from $-\sqrt{4- x^2}$ to $-\sqrt{1- x^2}$,z from $\sqrt{x^2+ y^2}$ to $\sqrt{4- x^2- y^2}$

The fourth with x from $\sqrt{2}/2$ to $\sqrt{2}$, y from $-\sqrt{4- x^2}$ to $\sqrt{4- x^2}$,z from $\sqrt{x^2+ y^2}$ to $\sqrt{4- x^2- y^2}$

That might make more sense to you if you draw the circles in the xy-plane and mark out each of the five regions above

To actually find the volume, it would be much easier to use spherical coordinates. In spherical coordinates, $\rho$ goes from 1 to 2, $\theta$ goes from 0 to $2\pi$, and $\phi$ goes from 0 to $2\pi$.