Question

What is the surface area of the solid created by revolving `f(x) =e^(4x)-x^2e^(2x) , x in [1,2]` around the x axis?

Answer 1

Given that

`f(x)=y=e^{4x}-x^2e^{2x}`

Volume of solid obtained by revolving the above curve about the x-axis between `x=1` & `x=2`

`V=\int_1^2y^2\ dx`

`=\int_1^2(e^{4x}-x^2e^{2x})^2\ dx`

`=\int_1^2(e^{8x}+x^4e^{4x}-2x^2e^{6x})\ dx`

`=\int_1^2(e^{8x}+x^4e^{4x}-2x^2e^{6x})\ dx`
`=[e^{8x}/8+e^{4x}(x^4/4-x^3/4+{3x^2}/16-{3x}/32+3/128)-e^{6x}(x^2/3-x/9+1/54)]_1^2`

`=934337.8548`

The surface area is given as

`\int_1^{2}2\pi y\sqrt{1+(y')^2}\ dx`

`=\int_1^{2}2\pi (e^{4x}-x^2e^{2x})\sqrt{1+(e^{4x}-2xe^{2x}(x+1))^2}\ dx`

Solving above integral is much more complicated which can't be solved by elementary method