Change in the order of integration in triple integrals

  • #1
WMDhamnekar
MHB
329
28
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?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,650
Please show your work.
 
  • #3
WMDhamnekar
MHB
329
28
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)##
 
Last edited:
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,650
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##.
 
  • Like
Likes WMDhamnekar and Delta2
  • #5
WMDhamnekar
MHB
329
28
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
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,650
First, do you understand why you cannot reasonably perform the integral over ##x##?
 
  • #7
WMDhamnekar
MHB
329
28
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
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,650
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.
 
  • Like
Likes WMDhamnekar and Delta2
  • #9
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
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).
 
  • Like
Likes WMDhamnekar
  • #10
WMDhamnekar
MHB
329
28
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?
 
Last edited:
  • #11
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
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.
 
  • Like
Likes WMDhamnekar
  • #12
WMDhamnekar
MHB
329
28
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.
 
  • #13
Prof B
66
37
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.
 

Suggested for: Change in the order of integration in triple integrals

Replies
34
Views
901
  • Last Post
Replies
2
Views
600
Replies
16
Views
480
  • Last Post
Replies
2
Views
425
  • Last Post
Replies
4
Views
956
Replies
4
Views
267
Replies
17
Views
1K
Replies
3
Views
244
Replies
13
Views
2K
Replies
3
Views
247
Top