Change in the order of integration in triple integrals

• WMDhamnekar
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}##

WMDhamnekar

MHB
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?

Delta2

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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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##.

WMDhamnekar and Delta2
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?

First, do you understand why you cannot reasonably perform the integral over ##x##?

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

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.

WMDhamnekar and Delta2
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).

WMDhamnekar
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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Delta2
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.

WMDhamnekar
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.

Delta2
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.

Delta2