Welcome to our community

Be a part of something great, join today!

Limits of volume integrals

GreenGoblin

Member
Feb 22, 2012
68
Help choose the limits of the following volume integrals:

1) V is the region bounded by the planes x=0,y=0,z=2 and the surface z=x^2 + y^2 lying the positive quadrant. I need the limits in terms of x first, then y then z AND z first, then y and then x. And also polar coordinates, x=rcost, y=rsint, z=z.

I am trying to convince myself

2) V is the region bounded by x+y+z=1, x=0, y=0, z=0.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Help choose the limits of the following volume integrals:

1) V is the region bounded by the planes x=0,y=0,z=2 and the surface z=x^2 + y^2 lying the positive quadrant. I need the limits in terms of x first, then y then z AND z first, then y and then x. And also polar coordinates, x=rcost, y=rsint, z=z.

I am trying to convince myself

2) V is the region bounded by x+y+z=1, x=0, y=0, z=0.
The region V is the volume above the parabaloid surface \(z=x^2+y^2\) and below \(z=2\) in the first quadrant. The projection of this region onto the x-y plane is that part of the circle \( x^2 + y^2=2 \) in the first quadrant.

So this is \(x\in (0,\sqrt{2})\) and \( y \in (0, \sqrt{2-x^2}) \) and \( z \in ( \sqrt{x^2+y^2} , 2) \)

CB
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
(I assume that by "x first, then y then z" you are referring to going from left to right. Of course, the integration is in the opposite order.)

"x first, then y then z" is pretty much the usual way to do an integral like this. Note that if we project into the xy-plane, the upper boundary projects to the circle $x^2+ y^2= 2$, a circle of radius $2$, centered at (0, 0) and in the first quadrant that is a quarter circle. In order to cover that entire figure, clearly x has to go from 0 to $\sqrt{2}$. Now, for each x, y must go from the x-axis, y= 0, up to the circle, $y= \sqrt{2- x^2}$. And, finally, for any given (x,y), z must go from the paraboloid, $z= x^2+ y^2$, to z= 2. The integral would be $\int_{x=0}^\sqrt{2}\int_{y= 0}^\sqrt{2- x^2}\int_{z= x^2+ y^2}^2 f(x,y,z)dzdydx$.

In the other order, z first, then y and then x, project to the yz- plane. There, of course, the paraboloid project to half of the parabola $z= y^2$. Now, z goes from 0 to 2 and, for each z, y goes from 0 to $\sqrt{z}$. Now, for each y and z, x goes from 0 to $\sqrt{2- y^2}$. The integral would be $\int_{z=0}^2\int_{y= 0}^\sqrt{z}\int_{x= 0}^\sqrt{2- y^2} f(x,y,z)dxdydz$.

For V is the region bounded by x+y+z=1, x=0, y=0, z=0, with the order "x first, then y then z", we project to the xy-plane which gives the triangle bounded by x= 0, y= 0, and x+ y= 1. x ranges from 0 to 1 and, for each x, y ranges from 0 to 1- x. For each (x, y), z ranges from 0 to 1- x- y. The integral would be $\int_{x= 0}^1\int_{y= 0}^{1- x}\int_{z= 0}^{1- x- y} f(x,y,z) dzdydx$. Can you get it for the order "z first, then y then x"?
 
Last edited: