# Triple Integral - Simple one

• May 17th 2010, 10:37 AM
WannaBe
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:
$x=1$, $x=2$, $x-y+1=0$,
$x-2y=2$, $x+y-z=0$ , $z=0$...

Write the integral $\int \int \int_{A} f(x,y,z) dxdydz$ 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:
$A=( (x,y,z)|(x,y) \in E, g^1(x,y)\leq z \leq g^2(x,y)$. Then if f is a continous function with 3 variables, continous in A, then:
$\int \int \int_{A} f(x,y,z) dxdydz = \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z)dz) dxdy$...

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...

• May 17th 2010, 11:40 AM
AllanCuz
Quote:

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:
$x=1$, $x=2$, $x-y+1=0$,
$x-2y=2$, $x+y-z=0$ , $z=0$...

Write the integral $\int \int \int_{A} f(x,y,z) dxdydz$ 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:
$A=( (x,y,z)|(x,y) \in E, g^1(x,y)\leq z \leq g^2(x,y)$. Then if f is a continous function with 3 variables, continous in A, then:
$\int \int \int_{A} f(x,y,z) dxdydz = \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z)dz) dxdy$...

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

$0 \le z \le x+ y$

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 $1 \le x \le 2$

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

Clearly the first equation is greater

Thus,

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

We get,

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

Now you need to re-write for

$dzdxdy$

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
• May 17th 2010, 12:03 PM
WannaBe
Quote:

Originally Posted by AllanCuz
Clearly our z is bounded by

$0 \le z \le x+ y$

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 $1 \le x \le 2$

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

Clearly the first equation is greater

Thus,

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

We get,

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

Now you need to re-write for

$dzdxdy$

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 $1 \le x \le 2$ and $\frac{x}{2} + 1 \le y \le x + 1$ then : $1.5 \le y \le 3$ ... Hence we have in the signs of the theorem I've quoted:
$E= ( (x,y) | 1 \le x \le 2 , 1.5 \le y \le 3 )$... And then the triple integral of the function f in A IS:
$\int \int_{E} ( \int_{0}^{x+y} f(x,y,z) dz) dxdy$
And this is all we need to do...

Am I right in this?

Thanks !
• May 17th 2010, 12:15 PM
11rdc11
Quote:

Originally Posted by WannaBe
Hey there...
z is clearly bounded indeed...But in a similar way we can show that if $1 \le x \le 2$ and $\frac{x}{2} + 1 \le y \le x + 1$ then : $1.5 \le y \le 3$ ... Hence we have in the signs of the theorem I've quoted:
$E= ( (x,y) | 1 \le x \le 2 , 1.5 \le y \le 3 )$... And then the triple integral of the function f in A IS:
$\int \int_{E} ( \int_{0}^{x+y} f(x,y,z) dz) dxdy$
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.
• May 17th 2010, 12:31 PM
WannaBe
Quote:

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:
$E=( (x,y) | 1 \le x \le 2 , 1+\frac{x}{2} \le y \le 1+x )$ ?
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