# Rewrite Triple integral

1. Nov 27, 2011

### violette

Appreciate any help,thanks!

Rewrite the integral:
∫0<x<1 ∫0<z<1-x2 ∫0<y<1-x dxdzdy

into this form: $\int$$\int$$\int$ dxdydz

How do I change the integrals?Can any kind souls teach me how to sketch the diagram?I cant visualise it >.<

2. Nov 27, 2011

### cragar

x=1 is plane , z=1-x^2 graph this in the zx plane and you get a parabola and then extend this straight out on the y axis, it will look like you took a piece of paper and curved it into the shape of a paraboloid. then graph y=1-x in the xy plane, then put it all together.

3. Nov 27, 2011

### violette

hi thanks for the reply =)

4. Nov 28, 2011

### jackmell

Your notation is ambiguous. As you nest the integrals, the dimension should decrease as you go from the inner integral to the outer one. Looks like you could mean:

$$\int_0^1\int_1^{1-x}\int_0^{1-x^2} dzdydx$$

so it's 3-2-1: the inner integral is from a surface f(y,x) to another surface g(y,x). That's 3D. The centerr one goes from a curve h(x) to another curve p(x). That's 2D, and the outer one goes from the point a to b. That's 1D.

Start by drawing the 3D axes on a sheet of paper with the inner variable point up, center one going into the paper, outer one going horizontally. Then label them. Can you draw a nice picture of just that for now? Ok then. Start with the outer integral and start drawing. First one is easy. just make two points at x=0 and x=1 on your plot. Now the center integral: that one goes from the function y=h(x)=0 to y=p(x)=1-x. Now can you draw in the x-y plane of your plot, those two functions? Now the hard part: Can you next draw, even pretty poorly, the two surfaces f(x,y)=0 and g(x,y)=1-x^2? Start by just drawing the curve for g(x,0)=1-x^2. Then since it's independent of y, draw another curve g(x,1/2)=1-x^2, another one at g(x,3/4)=1/x^2. Try and approach this plot you're drawing as if that's all you had to do and had to draw a nice one to turn in. That'll help with the analysis.

Now when you change the order, it's still 3-2-1 but the functions are different: the inner integral goes from one surface to another and are functions of the two remaing variables, the center integral goes from one curve to the next. These functions are functions of the outer variable, and the outer variable just goes from one point to the next.

Last edited: Nov 28, 2011
5. Nov 28, 2011

### violette

Ah yes I manage to get!
THANKS!!
=D