Change in the order of integration in triple integrals

In summary: Isn't this a little bit more complicated than the hint suggested?In summary, if we solve the L.H.S. of this equation, we get ##\frac{(b-a)^3}{6}## and if we solve R.H.S. of this equation, we get ##-\frac{2b^3-3ba^2 +a^3}{6}##
  • #1
WMDhamnekar
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Homework Statement
Show that ##\displaystyle\int_a^b\displaystyle\int_a^z \displaystyle\int_a^y f(x) dx dy dz = \displaystyle\int_a^b \frac{(b-x)^2}{2} f(x) dx ##

Hint: Think of how changing the order of integration in the triple integral changes the limits of integration.
Relevant Equations
No equation
If we solve the L.H.S. of this equation, we get ## \frac{(b-a)^3}{6}## and if we solve R.H.S. of this equation, we get ##-\frac{2b^3-3ba^2 +a^3}{6}##

So, how can we say, this equation is valid?

By the way, how can we use the hint given by the author here?
 
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  • #2
Please show your work.
 
  • #3
Solving L.H.S. of the equation ## \displaystyle\int_a^b\displaystyle\int_a^z\displaystyle\int_a^y f(x) dx dy dz = \displaystyle\int_a^b\displaystyle\int_a^z (y-a)f(x) dy dz##
## L.H.S. = \displaystyle\int_a^b \left( \frac{y^2}{2}-ay \big |_{y=a}^{y=z}\right) dz##

##L.H.S.= \displaystyle\int_a^b \frac{z^2}{2}-az -\frac{a^2}{2}+ a^2 dz ##
## L.H.S. =\frac{z^3}{6}-\frac{az^2}{2} -\frac{a^2z}{2} + a^2z \big|_{z=a}^{z=b}##

## L.H.S.= \frac{b^3 -3ab^2 +3a^2b -a^3}{6}= \frac{(b-a)^3}{6}f(x)##

In this way, if we solve the R.H.S. of the given equation, we get ##-\frac{2b^3 -3a^2b +a^3}{6}f(x)##
 
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  • #4
WMDhamnekar said:
Solving L.H.S. of the equation ## \displaystyle\int_a^b\displaystyle\int_a^z\displaystyle\int_a^y f(x) dx dy dz = \displaystyle\int_a^b\displaystyle\int_a^z (y-a)f(x) dy dz##
## L.H.S. = \displaystyle\int_a^b \left( \frac{y^2}{2}-ay \big |_{y=a}^{y=z}\right) dz##

##L.H.S.= \displaystyle\int_a^b \frac{z^2}{2}-az -\frac{a^2}{2}+ a^2 dz ##
## L.H.S. =\frac{z^3}{6}-\frac{az^2}{2} -\frac{a^2z}{2} + a^2z \big|_{z=a}^{z=b}##

## L.H.S.= \frac{b^3 -3ab^2 +3a^2b -a^3}{6}= \frac{(b-a)^3}{6}f(x)##

In this way, if we solve the R.H.S. of the given equation, we get ##-\frac{2b^3 -3a^2b +a^3}{6}##

Your first equality is wrong. You cannot integrate ##f(x)## with respect to ##x## and end up with ##x f(x)## and then not insert the boundaries into ##f(x)##. Then you arbitrarily remove ##f(x))## in the next step.

If you evaluate both expressions for ##f(x) = 1## then they both evaluate to ##(b-a)^3/6##.
 
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  • #5
Orodruin said:
Your first equality is wrong. You cannot integrate ##f(x)## with respect to ##x## and end up with ##x f(x)## and then not insert the boundaries into ##f(x)##. Then you arbitrarily remove ##f(x))## in the next step.

If you evaluate both expressions for ##f(x) = 1## then they both evaluate to ##(b-a)^3/6##.
Then , how to answer this question using the given hint?
 
  • #6
First, do you understand why you cannot reasonably perform the integral over ##x##?
 
  • #7
Orodruin said:
First, do you understand why you cannot reasonably perform the integral over ##x##?
Yes. Because I could not insert the upper and lower limits of integration into f(x) and get the final answer to integrate it w.r.t. y
 
  • #8
WMDhamnekar said:
Yes. Because I could not insert the upper and lower limits of integration into f(x) and get the final answer to integrate it w.r.t. y
No, because ##f(x)## is unknown, you do not know what the primitive function is and cannot perform the integral in the first place. The function ##x f(x)## is generally not a primitive function of ##f(x)##.

So, since you cannot integrate with respect to ##x##, you will need to change the order of integration to get rid of some of the integrals. Note that ##f(x)## is not a function of ##y## or ##z## so it can be taken out of those integrals. As alluded to by the hint, when you do a change of the order of integration, you need to consider how this affects the integration boundaries.
 
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  • #9
Don't even think to perform integration with respect to x, you shouldn't (since the result to prove doesn't perform it either) and you can not since function f(x) is totally unknown.

Instead you can perform easily integration with respect to y and z but you first must find out how the limits of integration for y and z change if you put y first and z second (or z first and y second, they are different in each case).
 
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  • #10
Delta2 said:
Don't even think to perform integration with respect to x, you shouldn't (since the result to prove doesn't perform it either) and you can not since function f(x) is totally unknown.

Instead you can perform easily integration with respect to y and z but you first must find out how the limits of integration for y and z change if you put y first and z second (or z first and y second, they are different in each case).
So, to answer this question, we have to integrate in this way ##\Rightarrow \displaystyle\int_a^bf(x)\displaystyle\int_x^b\displaystyle\int_x^z 1 dy dz dx= \displaystyle\int_a^y \frac{(b-x)^2}{2}f(x) dx ##

Isn't it?
 
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  • #11
WMDhamnekar said:
So, to answer this question, we have to integrate in this way ##\Rightarrow \displaystyle\int_a^yf(x)\displaystyle\int_x^b\displaystyle\int_x^z 1 dy dz dx= \displaystyle\int_a^y \frac{(b-x)^2}{2}f(x) dx ##

Isn't it?
You got it almost correct, you correctly find out the limits for y and z but the limits for x are not entirely correct, I hope it is a typo from you.
 
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  • #12
Delta2 said:
You got it almost correct, you correctly find out the limits for y and z but the limits for x are not entirely correct, I hope it is a typo from you.
Thanks for pointing out my mistake. I have corrected it now.
 
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  • #13
There is a general method to safely change order of integration or summation. First write all of the ranges as inequalities, i.e.,
$$ a \leq x \leq y $$
$$ a \leq y \leq z $$
$$ a \leq z \leq b $$
Combine them into a single chain of inequalities, i.e.,
$$ a \leq x \leq y \leq z \leq b $$
Break that apart. If you want to integrate or sum over x last then start by writing
$$ a \leq x \leq b $$
Then deal with y and z.
 
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1. What is the order of integration in a triple integral?

The order of integration in a triple integral refers to the order in which the variables are integrated. In a triple integral, there are three variables: x, y, and z. The order of integration determines which variable is integrated first, second, and third.

2. How does changing the order of integration in a triple integral affect the result?

Changing the order of integration in a triple integral can affect the result by changing the limits of integration and the order in which the variables are integrated. This can lead to different values for the integral, as well as different interpretations of the integral in terms of physical or geometric quantities.

3. What is the purpose of changing the order of integration in a triple integral?

Changing the order of integration in a triple integral can make the evaluation of the integral easier or more efficient. It can also provide different perspectives or interpretations of the integral, which can be useful in certain applications.

4. How do you determine the appropriate order of integration in a triple integral?

The appropriate order of integration in a triple integral depends on the shape and orientation of the region being integrated, as well as the function being integrated. Generally, it is helpful to visualize the region and consider which variable would be easiest to integrate first, and then proceed accordingly.

5. Can the order of integration be changed in any triple integral?

Yes, the order of integration can be changed in any triple integral. However, it is important to note that changing the order of integration can sometimes lead to a more complicated integral, so it is important to consider the purpose and potential benefits before making a change.

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