1. Triple Integral

$\displaystyle \int\int\int xy dV$ where 0 <= x <= 3 & 0 <= y <= x & 0 <= z <= x+y

Thought it'd be simple, but apparently I'm missing something because my answer isn't matching the back of the book.

Help's appreciated.

2. Hello,

The bounds for z contain x and y, so you'll have to integrate firstly in respect with z.

Then, the bounds for y contain x, so integrate in respect with y before x.

$\displaystyle \int_0^3 \int_0^x \int_0^{x+y} xy \ dz dy dx$

$\displaystyle \int_0^3 \int_0^x \left[xyz\right]_0^{x+y} dy dx$

$\displaystyle \int_0^3 \int_0^x xy(x+y) \ dy dx$

$\displaystyle \int_0^3 \int_0^x yx^2 \ dy dx + \int_0^3 \int_0^x xy^2 \ dy dx$

$\displaystyle \int_0^3 x^2 \left[\frac{y^2}{2}\right]_0^x dx + \int_0^3 x \left[\frac{y^3}{3}\right]_0^x dx$

$\displaystyle \int_0^3 x^2 \frac{x^2}{2} dx + \int_0^3 x \frac{x^3}{3} dx$

$\displaystyle \frac{1}{2} \int_0^3 x^4 dx + \frac 13 \int_0^3 x^4 dx$

$\displaystyle \frac 56 \int_0^3 x^4 dx$

$\displaystyle \frac 56 \left[\frac{x^5}{5}\right]_0^3$

$\displaystyle \frac 56 \times \frac{3^5}{5}$

$\displaystyle \frac{81}{2}$

3. Hmhm, you got the same answer I did...glad I'm not alone, at least. ^^

The answer in the back of the book is 64/15.