Triple Integration Engineering Maths, Btech first year

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

$${\int }_0^1{\int }_0^2{\int }_0^{3}x{y}^2{z}^{3}dxdydz$$ The rule is same as Double Integration $$\text{Inner Variable}⟶\text{Inner Limit}$$ $$\text{Outer Variable}⟶\text{Outer Limit}$$ 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: $$={\int }_0^1{\int }_0^2{\left[\frac{x^2}{2}\right]}_0^3{y}^2{z}^{3}dydz$$ $$=\frac{9}{2}{\int }_0^1{\int }_0^2{y}^2{z}^{3}dydz$$ $$=\frac{9}{2}{\int }_0^1{\left[\frac{y^3}{3}\right]}_0^2{z}^{3}dz$$ $$=\frac{9}{2}\times \frac{8}{3}{\int }_0^1{z}^{3}dz$$ $$=12{\left[\frac{z^4}{4}\right]}_0^1$$ $$=\boxed{{\displaystyle 3}}$$

Type-2 : Variable Limits

$${\int }_0^a{\int }_0^x{\int }_0^{x+y}{e}^{x+y+z}dxdydz$$ 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: $$={\int }_0^a{\int }_0^x{\left[e^{x+y+z}\right]}_0^{x+y}dxdy$$ $$={\int }_0^a{\int }_0^x[e^{x+y+(x+y)}-e^{x+y+0}]dxdy$$ $$={\int }_0^a{\int }_0^x[e^{2x+2y}-e^{x+y}]dxdy$$ $$={\int }_0^a{[\frac{e^{2x+2y}}{2}-\frac{e^{x+y}}{1}]}_0^{x}dx$$ $$={\int }_0^a[(\frac{e^{2x+2x}}{2}-\frac{e^{x+x}}{1})-(\frac{e^{2x+0}}{2}-\frac{e^{x+0}}{1})]dx$$ $$={\int }_0^a[(\frac{e^{4x}}{2}-e^{2x})-(\frac{e^{2x}}{2}-e^x)]dx$$ $$={\int }_0^a[\frac{e^{4x}}{2}-\frac{3e^{2x}}{2}+e^x]dx$$ $$={[\frac{e^{4x}}{8}-\frac{3e^{2x}}{4}+e^x]}_0^a$$ $$=\frac{e^{4a}}{8}-\frac{3e^{2a}}{4}+e^a-(\frac{1}{8}-\frac{3}{4}+1)$$ $$=\frac{e^{4a}-6e^{2a}+8e^a+3}{8}$$