Question

Find the surface area of the solid of revolution obtained by rotating the curve y=2x^3 from x=1 to x=6 about the x-axis?

Answer 1

`color(blue)[S_A=2piint_1^6(2x^3)*sqrt[1+(6x^2)^2]*dx=5.863*10^5]`

Explanation:

The surface area of the solid revolution around the `"x-axis"` is given by:

`color(red)[S_A=2piint_a^by*sqrt[1+(y')^2]*dx`

we will find the surface area bounded x=1 , x=6 and the x-axis

`S_A=2piint_1^6(2x^3)*sqrt[1+(6x^2)^2]*dx`

`S_A=2piint_1^6(2x^3)*sqrt[1+(36x^4)]*dx`

`S_A=(2pi)/72int_1^6(144x^3)*sqrt[1+(36x^4)]*dx`

`=pi/36[2/3(1+(36x^4))^(3/2)]_1^6=[(pi*(36*x^4+1)^(3/2))/54]_1^6`

`=[4*(46657^(3/2)/216-37^(3/2)/216)*pi]=[((46657^(3/2)-37^(3/2))*pi)/54]`

`=5.863*10^5`