# Triple Integral setup problem

1. Apr 10, 2005

### stunner5000pt

Triple Integral setup...

$$\int \int \int_{G} 6x (z+y^3) dx dy dz$$ G bounded by $$x = 0, \ x = y, \ z = y-y^2, \mbox{and} \ z=y^2 - y^3$$
x from 0 to1
y from 0 to x
z from z=y-y^2 to y^2 - y^3
and the integration order becomes dz dy dx
would this give the right answer?

what aboiut this one
$$\int \int \int_{G} xy + xz dx dy dz$$

G bounded by z = x, z=2-x, z = y^2
z goes from 2-y^2 to y^2
y goes from 2-x to x
x goes from 0 to 2
and the integration order to dz dy dx

I think the second one is wrong. Please do help!!

2. Apr 10, 2005

### whozum

Can you explain your reasoning for 1? Particularly the x and y. What exactly are you trying to do in these problems?

3. Apr 10, 2005

### stunner5000pt

the question is evaluate the integral over the given bounded region G

for the first one z goes like the y function, that s fine
for the y takes on a min value of 0 and max of x and
for the x the min value is zero but im not sure about the max value ..

Last edited: Apr 10, 2005
4. Apr 10, 2005

### whozum

Think of x=0 and x=y as planes, not as lines.

Also, x takes on a min value of 0, and a max value of y. You don't know much about y.

Try imagining the region G in your head, its bounded by the yz plane, and the plane y=x and the z function.

You want to find how far y and x go given your constraints. Solving the functions would not be a bad idea.

Last edited: Apr 10, 2005
5. Apr 10, 2005

### stunner5000pt

if i were to solve what am i solving for???
which functions would i use?

x=0, x= y, z = y - y^2 , z = y^2 - y^3
the intesection of which surfaces??

6. Apr 10, 2005

### whozum

7. Apr 10, 2005

### stunner5000pt

ok i got the y ranges from 0 to 1
now for the x part since x goes from 0 to y, x goes from 0 to 1
but y has to be a function of x so y goes from 0 to x?? and then x from 0 to 1?

Last edited: Apr 10, 2005
8. Apr 10, 2005

### whozum

Last edited: Apr 10, 2005