Triple Integral Proof

Question

Our professor gave us a question to try by ourselves at home, saying that $a_1 \leq x \leq a_2,\;b_1 \leq y \leq b_2,\;c_1 \leq z \leq c_2$ how would you show $$\iiint \alpha\;dxdydz = \alpha(a_2-a_1)(b_2-b_1)(c_2-c_1)$$ I solved it using iterated integrals but he said to attempt it without it and use the definition of a triple integral. I've been stumped at how to approach this, how would I start off?

Answer 1

Given positive integers $p,q,r$, we take partitions of the three intervals by putting $$ x_j=a_1+\frac{j(a_2-a_1)}{p},\ \ \ y_k=b_1+\frac{k(b_2-b_1)}{q},\ \ \ z_\ell=c_1+\frac{\ell(c_2-c_1)}{r}. $$ Then the Riemman sums for your integral are \begin{align} \sum_{j=1}^p\sum_{k=1}^q\sum_{\ell=1}^r\,f(x_j,y_k,z_\ell)\,\Delta x\,\Delta y\,\Delta z &=\sum_{j=1}^p\sum_{k=1}^q\sum_{\ell=1}^r\alpha\,\left(\frac{a_2-a_1}p\right)\,\left(\frac{b_2-b_1}q\right)\,\left(\frac{c_2-c_1}r\right)\\ \ \\ &=\alpha\,\left( {a_2-a_1}\right)\,\left( {b_2-b_1}\right)\,\left( {c_2-c_1}\right). \end{align} Then we need to take the limit as $p,q,r$ go to infinity, but the sums are already constant so the limit is the same.

Answer 2

There is no need for iterated integrals \begin{eqnarray*} \iiint \alpha\;dxdydz = \alpha \int_{a_1}^{a_2} dx \int_{b_1}^{b_2} dy \int_{c_1}^{c_2} dz = \cdots \end{eqnarray*}