Question

The value of the definite integral \(\displaystyle \int_{-3}^{3}\int_{-2}^{2}\int_{-1}^{1}\big(4x^{2}y - z^{3}\big)\,dz\,dy\,dx\) is ________. (Rounded off to the nearest integer)

Hint:

Before grinding through multivariable integrals, check for odd/even symmetry over symmetric limits. Odd parts vanish immediately, saving time.

Answer 1

Step 1: Integrate w.r.t. \(z\)
\(\displaystyle \int_{-1}^{1} \big(4x^{2}y - z^{3}\big)\,dz
= \left[4x^{2}y\,z - \frac{z^{4}}{4}\right]_{-1}^{1}
= 4x^{2}y(1-(-1)) - \frac{1-1}{4}
= 8x^{2}y.\)
\(\Rightarrow\) The triple integral reduces to \(\displaystyle \int_{-3}^{3}\int_{-2}^{2} 8x^{2}y\,dy\,dx.\)
Step 2: Integrate w.r.t. \(y\)
\(\displaystyle \int_{-2}^{2} 8x^{2}y\,dy = 8x^{2}\left[\frac{y^{2}}{2}\right]_{-2}^{2} = 4x^{2}\,(4-4)=0.\)
\(\Rightarrow\) The inner two integrals already give \(0\), so the outer \(x\)-integral is also \(0\).
(Equivalent symmetry argument)
\(4x^{2}y\) is odd in \(y\) over \([-2,2]\Rightarrow\) integrates to \(0\). \(-z^{3}\) is odd in \(z\) over \([-1,1]\Rightarrow\) integrates to \(0\). Hence the whole triple integral is \(0\). \[ \boxed{0} \]