# find surface area obtained by rotating `y=r^2,r in[1,4]`   about the y-axis

llltkl | Student

Sorry! I forgot to change the range of y in my earlier answer.

The surface area obtained by rotating a curve of a curve x=f(y) about the y-axis, in the interval [a,b], is given by the relation:

`S=int_a^b(2pixsqrt(1+((dx)/(dy))^2)dy)`

Here, `y=r^2`

`rArr r=sqrty`

`(dr)/(dy)=1/2y^(-1/2)`

Also, note that when r=1, y=1, and when r=4, y=16

So,

`S=int_1^16(2pi*rsqrt(1+(1/2y^(-1/2))^2dy)`

`= int_1^16(2pi*sqrty*sqrt(1+1/(4y)dy)`

`=pi int_1^16(sqrt(4y+1)dy)`

Let, 4y+1=t

4dy = dt, dt=1/4*dy

(When y=1, t=5 and when y=16, t=65)

therefore, `S= pi/4int_5^65sqrttdt`

`=pi/6(65sqrt65-5sqrt5)`

llltkl | Student

The surface area obtained by rotating a curve of a curve x=f(y) about the y-axis, in the interval [a,b], is given by the relation:

`S=int_a^b(2pixsqrt(1+((dx)/(dy))^2)dy)`

Here, `y=r^2`

`rArr r=sqrty`

`(dr)/(dy)=1/2y^(-1/2)`

So, `S=int_1^4(2pi*r*sqrt(1+(1/2y^(-1/2))^2)dy)`

`=int_1^4(2pi*sqrtysqrt(1+1/(4y))dy)`

`=int_1^4(2pi*sqrtysqrt((4y+1)/(4y))dy)`

`=piint_1^4(sqrt((4y+1))dy)`

Put `(4y+1)=t`

`4dy=dt`

`rArr dy=1/4dt`

(Note that when y=1, t=5 and when y=4, t=17)

`therefore S=pi/4int_5^17(sqrttdt)`

`=pi/6(17sqrt17-5sqrt5)`