Question

Evaluate :-

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red \int \limits_{0}^{3} \int \limits_{0}^{2} \int \limits_{0}^{1} ( x + y + z ) dx.dy.dz } } } { \bigstar} } }$

Topic :- Triple Integrals

Standard :- B.Sc ​

Answer 1

$\boxed{ \begin{array}{cc} \bf \to \: given : \\ \\ \rm \: \displaystyle \int \limits_{0}^{3} \int \limits_{0}^{2} \int \limits_{0}^{1} \bf \: ( x + y + z ) \: dx.dy.dz \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \int \limits_{0}^{2}\bigg \{ \int \limits_{0}^{1} \bf \: ( x + y + z ) dx\bigg \} \: .dy.dz \\ \\ \pink{ \sf \: integrate \: w.r.t \: \bf \: x} \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \int \limits_{0}^{2} \bf \: \bigg[\frac{ {x}^{2} }{2} + xy + xz\bigg]_{0}^{1}.dy.dz \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \int \limits_{0}^{2} \bf \: \bigg( \frac{ {1}^{2} }{2} + 1.y+1.z - 0\bigg)dy.dz \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \int \limits_{0}^{2} \bf \: \bigg(\frac{1}{2} + y + z\bigg )dy.dz \end{array}}$

$\boxed{ \begin{array}{cc} \pink { \sf \: integrate \: w.r.t \: \bf \: y} \\ \\\rm \: = \: \displaystyle \int \limits_{0}^{3} \bf\bigg \{ \int \limits_{0}^{2}\bigg (\frac{1}{2} + y + z\bigg)dy\bigg \} \: .dz \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \bf \: \bigg [\frac{y}{2} + \frac{ {y}^{2} }{2} + yz\bigg ]_{0}^{2} \: dz \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \bf \: \bigg( \frac{2}{2} + \frac{ {2}^{2} }{2} + 2z - 0\bigg) \: dz \\ \\ \rm \: = \: \displaystyle \int \limits_{0}^{3} \bf \: (1 + 2 + 2z) \: dz \\ \\\rm \: = \: \displaystyle \int \limits_{0}^{3} \bf \: (3 + 2z) \: dz \\ \\ \pink{ \sf \: integrate \: w.r.t \: \bf \: z} \\ \\ \rm \: = \: \bf\bigg[3z + \dfrac{2 {z}^{2} }{2} \bigg]_{0}^{3} \\ \\ \rm \: = \bf \: \bigg[3z + {z}^{2} \bigg]_{0}^{3} \\ \\ \rm \: = \: (3 \times 3 + {3}^{2}\ - 0)\ \\ \\ \: \rm \: = \: 9 + 9 \\ \\ \rm \: = \: 18 \\ \\ \: \end{array}}$

$\orange{\boxed{\boxed{\begin{array}{cc} \rm \: \therefore \: \displaystyle \int \limits_{0}^{3} \int \limits_{0}^{2} \int \limits_{0}^{1} \bf \: ( x + y + z ) \: dx.dy.dz = 18\end{array}}}}$