Question

Evaluate the double integral sin(y^3)dA where R is the region enclosed by y=√x , y=2,y=0 .

Answer 1

Given region is -

$f(x,y)=sin^3y$ and region is $y=\sqrt{x},y=0,y=2$

So The integral is given by

The limits for double integral for $x= 0 \text{ to }x=y^2$ and for $y=0 \text{ to } y=2$

=$\int_0^2\int_0^{y^2}sin(y^3)dxdy$

=$\int_0^2sin(y^3)[x]_0^{y^2}dy$

=$\int_0^2sin(y^3)y^2dy$

let $y^3=t\implies 3y^2dy=dt\implies y^2dy=\dfrac{dt}{3}$

At $y=0, t=0 \text { and at } y=2,t=y^3=(2)^3=8$

$=\int_0^{8}\dfrac{sint}{3}dt$

$=\dfrac{1}{3}[-cost]_0^{8}$

$=\dfrac{1}{3}(cos0-cos64)$

$=\dfrac{1}{3}(1-cos8)$

$=\dfrac{1}{3}(1-0.964)$

$\\=\dfrac{0.036}{3}\\=0.012 \text{ sq. units }$