# Triple Integral - Simple one

#### 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 $$\displaystyle g^1, g^2\(\displaystyle 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...

#### AllanCuz

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 $$\displaystyle g^1, g^2\(\displaystyle 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...

$$\displaystyle \(\displaystyle 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\)\)

#### WannaBe

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 !

#### 11rdc11

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.

#### WannaBe

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