What is the surface area of the solid created by revolving f(x) = x^3 , x in [1,2]f(x)=x3,x∈[1,2] around the x axis?

1 Answer
mason m
Feb 25, 2017

S=2piint_1^2x^3sqrt(1+9x^4)dx=pi/27(145^(3/2)-10^(3/2))approx199.4805S=2π∫21x3√1+9x4dx=π27(14532−1032)≈199.4805

Explanation:

The surface area of a solid created by a function revolved the xx-axis on x in[a,b]x∈[a,b] is:

S=2piint_a^bf(x)sqrt(1+(f'(x))^2)dx

Here we see that f(x)=x^3 and f'(x)=3x^2 so:

S=2piint_1^2x^3sqrt(1+9x^4)dx

We can integrate this. Let u=1+9x^4 so du=36x^3dx. Due to this, we need to multiply the interior of the integral by 36 and the exterior by 1//36.

S=(2pi)/36int_1^2 36x^3sqrt(1+9x^4)dx

Before going from x to u, we will have to transform the bounds. x=1 becomes u=1+9(1^4)=10. x=2 becomes u=1+9(2^4)=145.

S=pi/18int_10^145sqrtudu

S=pi/18int_10^145u^(1/2)du

S=pi/18[2/3u^(3/2)]_10^145

S=pi/18(2/3(145)^(3/2)-2/3(10)^(3/2))

S=pi/27(145^(3/2)-10^(3/2))approx199.4805

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