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

LCKurtz

This image shows the projection of the intersection of the two surfaces onto the yz plane. The equation of the cylindrical "wall" (shown in red) that projects onto the yz plane is gotten by setting the x values of the intersecting surfaces equal to each other. The black curve in the zy plane is where the wall intersects the plane and gives you the limits on the two zy integrals that have different "front" surfaces.

Fixed x and y were reversed.

Last edited: Jul 9, 2010