# Homework Help: Rewriting triple integrals

1. Jul 6, 2010

### EV33

1. The problem statement, all variables and given/known data
Rewrite this integral the other five ways
$$\int_{x=0}^{1}\int_{z=0}^{1-x^2}\int_{y= 0}^{1-x} dydzdx$$

2. Relevant equations
Must be in rectangular coordinates

3. The attempt at a solution
1.)$$\int_{z=0}^{1}\int_{x=0}^{\sqrt{1-z}}\int_{y= 0}^{1-x} dydxdz$$

2.)$$\int_{x=0}^{1}\int_{y=0}^{1-x}\int_{z=0}^{1-x^2}dzdydx$$

3.)$$\int_{y=0}^{1}\int_{x=0}^{1-y}\int_{z=0}^{1-x^2}dzdxdy$$

4.)$$\int_{z=0}^{1}\int_{y=0}^{1-x}\int_{x=0}^{\sqrt{1-z}}dxdydz$$

5.)$$\int_{y=0}^{1}\int_{z=0}^{1-x^2}\int_{x= 0}^{1-y}dxdzdy$$

With the last two I see the problem that the variable x will still be in the final answer. How can this problem be fixed. I figured the solution is to write the planes in another form but I don't see how to write the equations y=1/x or z=1-x2 without the variable x.

Last edited: Jul 6, 2010
2. Jul 7, 2010

### LCKurtz

The first step in solving a problem like this is to draw a picture of the 3d figure. Have you done that? You need it to see the proper limits.

The reason x first is tricky is that when you go from x on the back, which is x = 0, to x on the front, the front surface is in two pieces, partially a plane and partially a curved surface. You will need two integrals.

Last edited: Jul 7, 2010
3. Jul 9, 2010

### EV33

Yes I have drawn it. I am having trouble figuring out how to split it up. Could I by chance get a hint on how to split it up? Thank you.

4. Jul 9, 2010

### vela

Staff Emeritus
For #4, try sketching on the xy plane the intersection of the solid and a plane of constant z. It may be easier to ignore the constraint on y initially and then add it back in later.

Similarly, for #5, sketch on the xz plane the intersection of the solid and a plane of constant y. Again, it may be easier initially to ignore the constraint on x and then add it back in later.

5. Jul 9, 2010

### HallsofIvy

this tells that x lies between 0 and 1, for each x, z lies between 0 and $1- x^2$, and for each x and z, y lies between 0 and 1- x.

In the xz-plane, $z= 1- x^2$ is the parabola with vertex at (0, 1) and x-intercepts (1,0) and (-1,0). In the xy-plane, y= 1- x is the line through (0, 1) and (1, 0).

Notice that there is no "z" in that last! That tells us that we can write $\int\int\int dzdydx$ in exactly the same way.

For something like $\int\int\int dxdzdy$, we have to first think, what is the range for y? Looking at y= 1- x, we see that y can range from 0 to 1 as x goes between 0 and 1. The "outer" integral is $\int_0^1 dy$. Now, the range of z, for each y, is a bit more complicated. Since z goes from 0 up to $1- x^2$ and y from 0 to 1- x, we have x= 1- y and z from 0 to $1- (1-y)^2= 2y- y^2$. Finally, for all y and z, $z= 1- x^2$ is the same as $x^2= 1- z$ or $x= \pm\sqrt{1- z}$. x can go from $-\sqrt{1- z}$ to $\sqrt{1- z}$:
$$\int_{y=0}^1\int_{z= 0}^{2y- y^2}\int_{z=-\sqrt{1-z}^{\sqrt{1-x}}dxdzdy$$

Now, try the others.

6. Jul 9, 2010