Integration of 1 x 2 x 1 involves evaluating a triple integral over a rectangular prism whose edges measure one unit, two units, and one unit respectively. In three‑dimensional space this region can be described as the set of points ((x,y,z)) satisfying (0\le x\le 1), (0\le y\le 2), and (0\le z\
0 ≤ z ≤ 1. So to compute the volume, we evaluate the triple integral of 1 over this region. By Fubini’s theorem, we can decompose the integral into three successive single-variable integrals Small thing, real impact..
[ \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} 1 , dx
Continuing the computation, we now evaluate the innermost integral with respect to (x):
[ \int_{0}^{1} 1 , dx = \bigl[x\bigr]_{0}^{1}=1 . ]
With this result the double integral reduces to
[ \int_{0}^{2}\int_{0}^{1} 1 , dy , dz . ]
Performing the next integral,
[ \int_{0}^{2} 1 , dy = 2 , ]
so the remaining expression becomes
[ \int_{0}^{1} 2 , dz = 2\bigl[z\bigr]_{0}^{1}=2 . ]
Hence the triple integral of the constant function (1) over the given rectangular prism evaluates to (2). This confirms that the geometric interpretation—area (1\times 2 = 2) multiplied by the height (1)—is consistent with the analytic calculation. In practice, the application of Fubini’s theorem shows that integrating in the order (dx,dy,dz) indeed yields the correct volume without any loss of generality. So, the volume of the region defined by (0\le x\le1,;0\le y\le2,;0\le z\le1) is 2 cubic units.
Beyond the straightforward evaluation shown above, the same volume can be obtained by integrating in any order, a direct consequence of Fubini’s theorem for integrable functions on rectangular boxes. To give you an idea, choosing the order (dy,dz,dx) gives
[ \int_{0}^{1}!Day to day, \int_{0}^{1}! \int_{0}^{2} 1,dy,dz,dx =\int_{0}^{1}!\int_{0}^{1} 2,dz,dx =\int_{0}^{1} 2,dx =2 .
Similarly, the order (dz,dx,dy) yields
[ \int_{0}^{2}!\int_{0}^{1}!Now, \int_{0}^{1} 1,dz,dx,dy =\int_{0}^{2}! \int_{0}^{1} 1,dx,dy =\int_{0}^{2} 1,dy =2 .
Each permutation collapses the inner integrals to the length of the corresponding interval, and the product of the three side lengths—(1\times2\times1)—remains invariant. This invariance illustrates a broader principle: the volume of any axis‑aligned rectangular prism in (\mathbb{R}^3) equals the product of its side lengths, irrespective of the integration sequence chosen.
The computation also serves as a simple sanity check for more complex integrands. And if one replaces the constant integrand (1) by a function (f(x,y,z)) that separates as (f(x,y,z)=g(x)h(y)k(z)), the triple integral factorizes into the product of three one‑dimensional integrals, each evaluated over its respective interval. For non‑separable functions, the same limits apply, and the order of integration can still be rearranged to simplify the inner integrals—often a valuable technique in applied problems such as computing mass, charge, or probability over a rectangular domain.
Real talk — this step gets skipped all the time Not complicated — just consistent..
In practical settings, knowing that the volume of the unit‑by‑two‑by‑one box is exactly two cubic units allows quick scaling arguments. As an example, if a material with uniform density (\rho) fills this region, its mass is simply (m=\rho \times 2). Likewise, in probability theory, a joint density that is constant over this box must equal (1/2) to integrate to one, reflecting the reciprocal relationship between volume and uniform density That's the whole idea..
Thus, the triple‑integral approach not only confirms the elementary geometric formula but also reinforces the flexibility and power of iterated integration in multivariable calculus.
Conclusion: The volume of the rectangular prism defined by (0\le x\le1), (0\le y\le2), and (0\le z\le1) is unambiguously two cubic units, a result that holds regardless of the order of integration and serves as a foundational example for more layered volume and integral calculations.
