Question

How do you find the area of the surface generated by rotating the curve about the y-axis `y=1/3(x^2+2)^(3/2), 1<=x<=2`?

Answer 1

The surface area of a curve `y` rotated around the `x`-axis on `alt=xlt=b` is given by:

`S=2piint_a^bysqrt(1+(y')^2)dx`

Where `y=1/3(x^2+2)^(3/2)` we see that `y'=1/3(3/2(x^2+2)^(1/2))(2x)=xsqrt(x^2+2)`.

Then the surface area is:

`S=2piint_1^2 1/3(x^2+2)^(3/2)sqrt(1+(xsqrt(x^2+2))^2)dx`

`S=(2pi)/3int_1^2(x^2+2)^(3/2)sqrt(1+x^2(x^2+2))dx`

Note that `sqrt(1+x^2(x^2+2))=sqrt(x^2+2x^2+1)=sqrt((x^2+1)^2)=x^2+1`:

`S=(2pi)/3int_1^2(x^2+2)^(3/2)(x^2+1)dxapprox69.9055`