# Triple Integral Evaluation

1. Dec 6, 2006

### mathzeroh

Triple Integral Evaluation (quick and easy)

1. The problem statement, all variables and given/known data
$\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{3y} ({y+2x^2z})dz dy dx$

2. Relevant equations

None.

3. The attempt at a solution

Here is what I got at the end (the LaTeX takes too long to code in here, plus its not showing up):

59/36 because after integrating the whole thing, and then putting in the very last limits (0 and 1), all of the x's go away leaving just the coefficients which I worked out to get 59/36...can someone please verify this for me? i've checked it twice and got the same solution..it takes about 3 minutes to do if you're a genious (unlike me, so i'm trying to appeal to the math geniuses)..

thanks!

Last edited: Dec 6, 2006
2. Dec 7, 2006

### dextercioby

I get (actually Maple) $\frac{32}{21}$

∫{0}^{3y}(y+2x²z)dz= 9x²y²+3y²

∫_{x²}^{1}( 9x²y²+3y²)dy= -3x^{8}-x^{6}+3x²+1

∫_{0}^{1}(-3x^{8}-x^{6}+3x²+1) dx= ((32)/(21))

3. Dec 7, 2006

did it by hand and got 32/21

4. Dec 8, 2006

### dextercioby

I did it by hand, too. :tongue2: I had to type everything and then use the "ctrl c-ctrl v" keys. It's not as easy as it seems...

Daniel.

5. Dec 8, 2006

### HallsofIvy

$$\int_0^1\int_{x^2}^1\int_0^3y (y+ 2x^2z)dzdydx$$
$$\int_0^1\int_{x^2}^1\left[yz+ x^2z^2\right]_0^{3y}dydx$$
$$3\int_0^1\int_{x^2}^1(1+ x^2)y^2 dy dx$$
$$\int_0^1\left[(1+3x^2)y^3\right]_{x^2}^1 dx$$
$$\int_0^1(1+3x^2)(1-x^6)dx$$
$$\int_0^1(-3x^8- x^6+ 3x^2+ 1)dx$$
$$\left[-\frac{1}{3}x^9-\frac{1}{7}x^7+ x^3+ x\right]_0^1$$
$$-\frac{1}{3}-\frac{1}{7}+ 1+ 1$$
$$-\frac{7}{21}-\frac{3}{21}+\frac{21}{21}+\frac{21}{21}$$
$$\frac{32}{21}$$