# Multiple Integration

1. May 12, 2006

### AlmostFamous

This is a question i'm having trouble with.

Solve this equation for a

$$\iiint \,dz\,dy\,dx$$ = 2/7

I'm new to this code so i'll write the limits here.

the first integral sign is between 0 and 1
the second is between 0 and 4-a-x^2
the third is between 0 and 4-x^2-y

I've got to the third integral but this is what i've got to integrate

16-4a-8x^2+ax^2+x^4 - (16-8a-8x^2+2ax^2+a^2+x^4)

The bracketed part is all over 2, but again i don't know the code, sorry.

how would i go further with this? It's the bit where the bracket is over 2 that i'm having trouble with. Do i get rid of the brackets somehow or try to integrate the next part as it is? Any help would be much appreciated.

EDIT: I took the integral code from the LaTex thread, but the order should read dzdydx

2. May 12, 2006

### marlon

Let's go ove this step by step. After the first integration to z , you get 4-x^2-y, right ? This is the integrandum for the second integration to y. Perform this integration and fill in the boundaries. Do not calculate any squares, just leave the brackets as they are.

marlon

3. May 12, 2006

### HallsofIvy

Staff Emeritus
Click on the following to see the code:
(or click on the "quote" button to quote it in raw form)
$$\int_0^1\int_0^{4-a-x^2}\int_0^{4-x^2-y}dzdydx$$
The first integral is, of course, easy:
$$\int_0^1\int_0^{4-a-x^2}(4-x^2-y)dydx$$
$$\int_0^1\left(4y-x^2y-\frac{1}{2}y^2)\right)_0^{4-a-x^2}dx[/itex] [tex]\int_0^1\left(16-4a-4x^2-4x^2+ax^2+x^4-\frac{1}{2}(16-2a+a^2-8x^2+2ax^2+ x^4)\right)dx[/itex] which is what you have. Of course, you "get rid of the brackets" by multiplying each term by 1/2. Now combine like terms of x: The constant terms are [tex]16-4a- 8+ 4a- \frac{1}{2}a^2= 8- \frac{1}{2}a^2[/itex] The "x2" terms are [tex](-8+ a+ 4- a)x^2= -4x^2$$
The "x4" terms are
$$(1-\frac{1}{2})x^4= \frac{1}{2}x^4[/itex] so the integral is [tex]\int_0^1\left(8- \frac{1}{2}a^2-4x^2+\frac{1}{2}x^4\right)dx$$

It should be easy to integrate this. Once you do, you'll have a quadratic equation to solve for a.

Last edited: May 12, 2006
4. May 12, 2006

### AlmostFamous

Thats the way I did it at first, but I got a much different answer. I checked my original work and I got a sign the wrong way and the answer turned out to be rubbish. Got it sorted now though. Thanks alot.

Last edited: May 12, 2006