# Thread: Help with finding Boundaries for Triple INtegrals

1. ## Help with finding Boundaries for Triple INtegrals

For example:

x^2 + y^2 + z^2 <= R^2

z >= 0

Can someone help me find the x,y, and z boundaries? Would x and y be -R and R? And would z be 0 and R^2 - x^2 -y^2?

Also, for:

x^2 + y^2 <= R^2 and 0<= z <= H

I'm confused on the boundaries for this one as well. Anyone?

Last one;

If the question only says that a cub with edge length a, what would be the x,y,z boundaries for the triple integral?

2. Here you will need strong visualization skills. The inequality $x^2+y^2+z^2\leq R^2$ describes a solid sphere with radius $R$ centered on the origin. If $z\geq 0$, then we are only talking about the top half of the figure, the part above the xy-plane, a "semi-sphere". This shape is the domain $D=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2\leq R^2\}$ for the function $r\to\mathbb{R}" alt="r\to\mathbb{R}" />, $r(x,y,z)$ not defined in your problem. So we are actually finding the hypervolume of a 4-dimensional shape.

The bounds on $z$ go from $0$ to $R$, as given in the problem, so that will be the innermost integral. So we are taking horizontal cross-sections of the semi-sphere, giving us circular disks of radius $\sqrt{R^2-z^2}$. So $y$ must range from $-\sqrt{R^2-z^2}$ to $+\sqrt{R^2-z^2}$. The intersection of these two planes give a line of length $2\sqrt{R^2-z^2-y^2}$. Therefore the outermost integral will be $x$ ranging from $-\sqrt{R^2-z^2-y^2}$ to $+\sqrt{R^2-z^2-y^2}$. Hence,

$I=\int_{-\sqrt{R^2-z^2-y^2}}^{+\sqrt{R^2-z^2-y^2}}\int_{-\sqrt{R^2-z^2}}^{+\sqrt{R^2-z^2}}\int_{0}^{R} r(x,y,z) dzdydx$

(I suppose it would be inappropriate in this particular problem to convert to polar or cylindrical coordinates, which would incredibly simplify the problem.)

3. I think that most (in cartesian coords although spherical polars is the way to go) would put

$
\int_{_R}^{R} \int_{ -\sqrt{R^2-x^2} }^{\sqrt{R^2-x^2}} \int_0^{\sqrt{R^2-x^2-y^2}} f(x,y,z)dz\,dy\,dx
$

@media man - I think you want to reverse the order of your integrals.