Is there a simpler way to integrate this?

aruwin

Hi, I was wondering if there's a simpler way to integrate this? I got the answer by expanding one by one but that's such a long process! I got the answer 104/5 which is correct,though.


∫∫∫D (x + y + z)^4 dxdydz
D = {(x,y,z) | -1≤ x ≤1, -1≤y≤1, -1≤z≤ }

chiro

Hey aruwin.

For your integral, you are dealing with a simple rectangular volume, so the integral in its form is best evaluated by doing each variable separately at a time.

If you did it this way, then there's no point stressing about a better way to do it if you know and understand how to evaluate the integral.

genericusrnme

you could keep in your head that the integral if (x+a)^n = (1/n)(x+a)^(n+1) if you wanted but there's nothing wrong with expanding everything out

aruwin

How do I do each variable sperately at a time? You mean like ∫dx∫dy∫dz ?

Ok,what about the function inside it? When say rectangle, it means
(x+y+z)²(x+y+z)²,right? But still,how do I integrate it like this?

chiro

You do it in the form of (x+a)^n with respect to dx and integrate out the x value. Then do the same for the y and the same for the z until you get a numerical quantity.

Since x,y,z are all orthogonal variables, you can do this with no problem just like you would integrated say xydydx by keeping one variable 'constant' and then integrating it with respect to the other.

aruwin

OK,I am gonna ask you something first, is this correct?

∫∫ (x+y+z)⁵/5 dydz

But what happens to the x inside the bracket?Shouldn't I do something about it?

genericusrnme

take [itex]\int \int \int (x+y+z)^4 dx dy dz[/itex] and stick some brackets in there [itex]\int \int ( \int (x+y+z)^4 dx ) dy dz[/itex]
Substitute y+z = a
perform integration
substitute a = y + z
remove brackets
repeat

aruwin

Won't that change the range for dx?

genericusrnme

nope, all you've done is substitute a for y + z as to avoid confusion, you've not touched x or dx at all (you haven't really altered anything)

chiro

It only changes the interval if the interval depends on those other variables.

To understand this, you need to understand orthogonality: things are orthogonal if changing one variable doesn't change another. We can write this in terms of dA/dB = 0 and dB/dA = 0 where neither change in accordance with the other variable.

If your limits were say from 0 to y^2 - z, then yes the limit would have to factor this in but because each variable is truly orthogonal, then this doesn't happen (in this case).

aruwin

So I did what you told me but I didn't get the answer :( Tell me what is wrong here.
Check,please....

chiro

You wrote in your final step 2⁷ instead of 2x1⁷

aruwin

Even so, I won't get the answer that is 104/5.

chiro

You made a major sign error in your integral:instead of + (-2 + z)^6 you wrote - (-2 + z)^6.

aruwin

I will try again.But other than that,I'm doing this the right method,arent I?

chiro

Yes. I checked the answer with your calculation using a calculator: it gave 20.8 which is 104/5.

aruwin

I calculated another 2 times but I didn't get the answer :( Are you sure my working there had only one mistake?