I'm working on arc length calculation and area of surface of revolution in calculus and I'm really quite stuck on the process of how to do this. Here is a particular problem that I'm struggling with:
Find the surface area of the surface of revolution generated by revolving the graph $$y=x^3; \qquad 0 \leq x \leq10$$ around the $x$-axis.
I've gone through the steps that I've learned to do (listed below) and the steps for the most part seem to make sense, however I keep ending up with incorrect answers. Please help! Below I listed my general process of approaching the problem.
$$\begin{align} &y = x^3 \\&y' = 3x^2 \\&(y')^2 = 9x^4\\&1 + (dy/dx)^2 = 1 + 9x^4\end{align}$$
Using the formula:
$$2\pi y \cdot\int(1 + (dy/dx)^2)^{1/2} dx$$
Here's how I set up the integral for the problem:
$$2\pi x^3\cdot \int_0^{10}(1+9x^4)^{1/2} dx.$$
This came out to $$2\pi x^3\cdot(6/5)\cdot(1+9(10^4))^{3/2}\cdot x^5$$ My final answer was $$2.035785969E16.$$
Please help me understand where I'm going wrong!