Triple Integration
In Triple Integration, we integrate w.r.t. 3 variables. Here we are going to study two types:
Type-1 : Constant limits
Type-2 : Variable limits
Type-1 : Constant limits
\[ \int_0^1 \int_0^2 \int_0^3 xy^2z^3dxdydz \] The rule is same as Double Integration \[ \text{Inner Variable} \longrightarrow \text{Inner Limit} \] \[ \text{Outer Variable} \longrightarrow \text{Outer Limit} \] So here \(x\) will have limits 0 to 3, \(y\) will have limits 0 to 2 and \(z\) will have limits 0 to 1. Let's integrate: \[ = \int^1_0 \int^2_0 \left[\frac{x^2}{2} \right]^3_0y^2z^3dydz \] \[ = \frac{9}{2} \int^1_0 \int^2_0 y^2z^3dydz \] \[ = \frac{9}{2} \int_0^1 \left[\frac{y^3}{3} \right]^2_0 z^3dz \] \[ = \frac{9}{2}\times \frac{8}{3} \int^1_0 z^3dz \] \[ = 12 \left[\frac{z^4}{4} \right]_0^1 \] \[ = \boxed{3} \]
Type-2 : Variable Limits
\[ \int^a_0 \int^x_0 \int^{x+y}_0 e^{x+y+z}dxdydz \] Like in Double Integration, here \(x + y\) is in terms of \(z\) so it will be limits of \(z\) and the limit \(x\) belongs to \(y\). Let's integrate: \[ = \int^a_0 \int^x_0\left[e^{x+y+z} \right]^{x+y}_0dxdy \] \[ = \int^a_0 \int^x_0\left[e^{x+y+(x+y)} - e^{x+y+0} \right]dxdy \] \[ = \int^a_0 \int^x_0\left[e^{2x+2y} - e^{x+y} \right]dxdy \] \[ = \int^a_0 \left[\frac{e^{2x+2y}}{2} - \frac{e^{x+y}}{1} \right]^x_0 dx \] \[ = \int^a_0 \left[\left(\frac{e^{2x+2x}}{2} - \frac{e^{x+x}}{1}\right) - \left(\frac{e^{2x+0}}{2} - \frac{e^{x+0}}{1} \right) \right]dx \] \[ = \int^a_0 \left[\left(\frac{e^{4x}}{2} - e^{2x}\right) - \left(\frac{e^{2x}}{2} - e^{x} \right) \right]dx \] \[ = \int^a_0 \left[\frac{e^{4x}}{2} - \frac{3e^{2x}}{2} + e^x \right]dx \] \[ = \left[\frac{e^{4x}}{8} - \frac{3e^{2x}}{4} + e^x \right]^a_0 \] \[ = \frac{e^{4a}}{8} - \frac{3e^{2a}}{4} + e^a - \left(\frac{1}{8} - \frac{3}{4} + 1 \right) \] \[ = \frac{e^{4a} - 6e^{2a} + 8e^a + 3}{8} \]
