Question

$\displaystyle \sf \red{ \int_{0}^{3} \int^{ \sqrt{9 - {x}^{2}}}_{ - \sqrt{9 - {x}^{2} } }( {x}^{3} + x {y}^{2} ) \: dydx}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given double integral is

$\rm :\longmapsto\:\displaystyle \sf { \int_{0}^{3} \int^{ \sqrt{9 - {x}^{2}}}_{ - \sqrt{9 - {x}^{2} } }( {x}^{3} + x {y}^{2} ) \: dydx}$

can be rewritten as

$\rm :\longmapsto\:\displaystyle \sf { \int_{0}^{3}\bigg[ \int^{ \sqrt{9 - {x}^{2}}}_{ - \sqrt{9 - {x}^{2} } }( {x}^{3} + x {y}^{2} ) \: dy\bigg]dx}$

Now, we have to first integrate with respect to y treating x as constant.

So,

$\red{\rm :\longmapsto\:f(y) = {x}^{3} + {xy}^{2}}$

So,

$\red{\rm :\longmapsto\:f( - y) = {x}^{3} + {x( - y)}^{2}= {x}^{3} + {xy}^{2} = f(y)}$

$\red{\rm \implies\:f(y) \: is \: even \: function}$

So, above integral can be rewritten as

$\rm \:  =  \: 2\displaystyle \sf { \int_{0}^{3} \bigg[\int^{ \sqrt{9 - {x}^{2}}}_{ 0}}( {x}^{3} + {xy}^{2})dy \: \bigg] \: dx$

$\rm \:  =  \: 2\displaystyle \sf { \int_{0}^{3} }\bigg[ {x}^{3}y + \dfrac{ {xy}^{3} }{3} \bigg] ^{ \sqrt{9 - {x}^{2}}}_{ 0}\: dx$

$\rm \:  =  \: 2\displaystyle\int_0^{3}\bigg( {x}^{3} \sqrt{9 - {x}^{2} } + \dfrac{x {( \sqrt{9 - {x}^{2} } )}^{3} }{3} \bigg) dx$

can be rewritten as

$\rm \:  =  \: \dfrac{2}{3} \displaystyle\int_0^{3}\bigg( 3x.{x}^{2} \sqrt{9 - {x}^{2} } + x {( \sqrt{9 - {x}^{2} } )}^{3}\bigg) dx$

$\rm \:  =  \: \dfrac{2}{3} \displaystyle\int_0^{3}\bigg( 3{x}^{2} + 9 - {x}^{2} \bigg)x \sqrt{9 - {x}^{2} } dx$

$\rm \:  =  \: \dfrac{2}{3} \displaystyle\int_0^{3}\bigg( 2{x}^{2} + 9 \bigg)x \sqrt{9 - {x}^{2} } dx$

$\rm \:  =  \: \dfrac{2}{3} \displaystyle\int_0^{3}\bigg( 2{x}^{2} + 9 \bigg)\sqrt{9 - {x}^{2} } \: xdx$

To evaluate this integral, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\: \sqrt{9 - {x}^{2} } = y}$

$\red{\rm :\longmapsto\:9 - {x}^{2} = {y}^{2}}$

$\red{\rm :\longmapsto\: - 2x \: dx \: = \: 2y \: dy}$

$\red{\rm :\longmapsto\: x \: dx \: = \: - \: y \: dy}$

So, on substituting these values, the above integral can be rewritten as

$\rm \:  =  \: \dfrac{2}{3} \displaystyle\int_3^{0}\bigg( 2(9 - {y}^{2}) + 9 \bigg)y \: ( - y)dy$

$\rm \:  =  \: - \dfrac{2}{3} \displaystyle\int_3^{0}\bigg( 18- 2{y}^{2}+ 9 \bigg) {y}^{2} dy$

$\rm \:  =  \: - \dfrac{2}{3} \displaystyle\int_3^{0}\bigg( 27 - 2{y}^{2} \bigg) {y}^{2} dy$

$\rm \:  =  \: - \dfrac{2}{3} \displaystyle\int_3^{0}\bigg( 27 {y}^{2} - 2{y}^{4} \bigg) dy$

$\rm \:  =  \: \dfrac{2}{3} \displaystyle\int_0^{3}\bigg( 27 {y}^{2} - 2{y}^{4} \bigg) dy$

$\rm \:  =  \: \dfrac{2}{3}\bigg[\dfrac{27 {y}^{3} }{3} - \dfrac{ {2y}^{5} }{5} \bigg]_0^{3}$

$\rm \:  =  \: \dfrac{2}{3}\bigg[9 {y}^{3} - \dfrac{ {2y}^{5} }{5} \bigg]_0^{3}$

$\rm \:  =  \: \dfrac{2}{3}\bigg[9 (27) - \dfrac{2 \times 243 }{5} \bigg]$

$\rm \:  =  \: \dfrac{2}{3}\bigg[243 - \dfrac{2 \times 243 }{5} \bigg]$

$\rm \:  =  \: \dfrac{2}{3} \times 243\bigg[1- \dfrac{2 }{5} \bigg]$

$\rm \:  =  \: \dfrac{2}{3} \times 243\bigg[\dfrac{5 - 2 }{5} \bigg]$

$\rm \:  =  \: \dfrac{2}{3} \times 243\bigg[\dfrac{3 }{5} \bigg]$

$\rm \:  =  \: \dfrac{486}{5}$