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Math Help - Triple Integral - Simple one

  1. #1
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    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...

    Thanks in advance
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WannaBe View Post
    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...

    Thanks in advance
    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
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  3. #3
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    Quote Originally Posted by AllanCuz View Post
    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 !
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by WannaBe View Post
    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.
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    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
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