# Thread: Triple Integral - Simple one

1. ## Triple Integral - Simple one

Hey there, I'll be delighted to get some help in the following question:
Let A be the region in space bouded by the next planes:
$\displaystyle x=1$, $\displaystyle x=2$, $\displaystyle x-y+1=0$,
$\displaystyle x-2y=2$, $\displaystyle x+y-z=0$ , $\displaystyle z=0$...

Write the integral $\displaystyle \int \int \int_{A} f(x,y,z) dxdydz$ as shown in the next theorem:
Let E be a closed region with a surface in R^2 and let [tex] g^1, g^2[tex] be two real functions, continous in E. Let's look at A:
$\displaystyle A=( (x,y,z)|(x,y) \in E, g^1(x,y)\leq z \leq g^2(x,y)$. Then if f is a continous function with 3 variables, continous in A, then:
$\displaystyle \int \int \int_{A} f(x,y,z) dxdydz = \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z)dz) dxdy$...

The problem is I can't figure out how the region A looks like...
Hope you'll be able to help me dealing with this question...

2. Originally Posted by WannaBe
Hey there, I'll be delighted to get some help in the following question:
Let A be the region in space bouded by the next planes:
$\displaystyle x=1$, $\displaystyle x=2$, $\displaystyle x-y+1=0$,
$\displaystyle x-2y=2$, $\displaystyle x+y-z=0$ , $\displaystyle z=0$...

Write the integral $\displaystyle \int \int \int_{A} f(x,y,z) dxdydz$ as shown in the next theorem:
Let E be a closed region with a surface in R^2 and let [tex] g^1, g^2[tex] be two real functions, continous in E. Let's look at A:
$\displaystyle A=( (x,y,z)|(x,y) \in E, g^1(x,y)\leq z \leq g^2(x,y)$. Then if f is a continous function with 3 variables, continous in A, then:
$\displaystyle \int \int \int_{A} f(x,y,z) dxdydz = \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z)dz) dxdy$...

The problem is I can't figure out how the region A looks like...
Hope you'll be able to help me dealing with this question...

Clearly our z is bounded by

$\displaystyle 0 \le z \le x+ y$

Since the bounds of x are given we will save those for last (keeping with the definition of the triple integral)

So all we need to find is our y bounds and we have 2 equations of y in terms of x, no problem!

We need to determine which of the following is greater on the interval $\displaystyle 1 \le x \le 2$

$\displaystyle y = x + 1$ or $\displaystyle \frac{x}{2} + 1$

Clearly the first equation is greater

Thus,

$\displaystyle \frac{x}{2} + 1 \le y \le x + 1$

We get,

$\displaystyle \int_1^2 dx \int_{ \frac{x}{2} + 1}^{ x + 1 } dy \int_0^{x+y} dz$

Now you need to re-write for

$\displaystyle dzdxdy$

Clearly we need to find constant values for z and get x/y in terms of z which should be no problem. Compute both and compare

3. Originally Posted by AllanCuz
Clearly our z is bounded by

$\displaystyle 0 \le z \le x+ y$

Since the bounds of x are given we will save those for last (keeping with the definition of the triple integral)

So all we need to find is our y bounds and we have 2 equations of y in terms of x, no problem!

We need to determine which of the following is greater on the interval $\displaystyle 1 \le x \le 2$

$\displaystyle y = x + 1$ or $\displaystyle \frac{x}{2} + 1$

Clearly the first equation is greater

Thus,

$\displaystyle \frac{x}{2} + 1 \le y \le x + 1$

We get,

$\displaystyle \int_1^2 dx \int_{ \frac{x}{2} + 1}^{ x + 1 } dy \int_0^{x+y} dz$

Now you need to re-write for

$\displaystyle dzdxdy$

Clearly we need to find constant values for z and get x/y in terms of z which should be no problem. Compute both and compare

Hey there...
z is clearly bounded indeed...But in a similar way we can show that if $\displaystyle 1 \le x \le 2$ and $\displaystyle \frac{x}{2} + 1 \le y \le x + 1$ then : $\displaystyle 1.5 \le y \le 3$ ... Hence we have in the signs of the theorem I've quoted:
$\displaystyle E= ( (x,y) | 1 \le x \le 2 , 1.5 \le y \le 3 )$... And then the triple integral of the function f in A IS:
$\displaystyle \int \int_{E} ( \int_{0}^{x+y} f(x,y,z) dz) dxdy$
And this is all we need to do...

Am I right in this?

Thanks !

4. Originally Posted by WannaBe
Hey there...
z is clearly bounded indeed...But in a similar way we can show that if $\displaystyle 1 \le x \le 2$ and $\displaystyle \frac{x}{2} + 1 \le y \le x + 1$ then : $\displaystyle 1.5 \le y \le 3$ ... Hence we have in the signs of the theorem I've quoted:
$\displaystyle E= ( (x,y) | 1 \le x \le 2 , 1.5 \le y \le 3 )$... And then the triple integral of the function f in A IS:
$\displaystyle \int \int_{E} ( \int_{0}^{x+y} f(x,y,z) dz) dxdy$
And this is all we need to do...

Am I right in this?

Thanks !
Nope do it the way AllanCuz told you. If you make your y bounds just numbers than your saying the region in the z=0 plane is a just a rectangle which it is not.

5. Originally Posted by 11rdc11
Nope do it the way AllanCuz told you. If you make your y bounds just numbers than your saying the region in the z=0 plane is a just a rectangle which it is not.

So the region is actually:
$\displaystyle E=( (x,y) | 1 \le x \le 2 , 1+\frac{x}{2} \le y \le 1+x )$ ?
The given order is to write the specific integral as shown in the theorem...No calculation is needed... Does my representation above is now correct? Or am I still wrong at something?

Hope you'll be able to help me understand my mistakes

Thanks