Find the area of the surface obtained by rotating the curve $y=\cosh(4x)/4$, $-1\leqslant x\leqslant 1$ about the $x$-axis.
I setup the integral as: $$2\pi\int_{-1}^{1}x\cdot \cosh(4x)/4dx$$
and ended up getting $0$ as the answer. Is that really the answer?
I also need to know how would one go about rotating this about the y-axis, but have no idea where to start.
Thanks.
You should review the formula for the surface area in the case of a surface of revolution (e.g. here).
The surface area of the surface obtained by rotation the graph of $y=f(x)$ about the $x$-axis on the interval $[x_1,x_2]$, is given by: $$2\pi\int_{x_1}^{x_2} y \sqrt{1+ \left(y'\right)^2}\,\mbox{d}x = 2\pi\int_{x_1}^{x_2} f(x) \sqrt{1+ \left(f'(x)\right)^2}\,\mbox{d}x$$ Now if $f(x) = \tfrac{\cosh(4x)}{4}$, then $f'(x)=\sinh(4x)$ so rotation on $[-1,1]$ gives: $$\frac{\pi}{2}\int_{-1}^{1} \cosh(4x) \sqrt{1+ \sinh^2(4x)}\,\mbox{d}x$$ You can simplify (a lot). Can you take it from here?
I also need to know how would one go about rotating this about the y-axis, but have no idea where to start.
The link from above also covers the formula for rotation about the $y$-axis.
You can get the surface areas due to rotation about either the $x$ or $y$ axes using Pappus's $(1^{st})$ Centroid Theorem, which states that the surface area A of a surface of revolution generated by rotating a plane curve C about an axis external to C and on the same plane is equal to the product of the arc length s of C and the distance d traveled by its geometric centroid (Pappus's centroid theorem). Simply put, $S=2\pi RL$, where R is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by
$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$
Thus we can say for your cases that
$$ S=2\pi\int_{-1}^1 y\sqrt{1+(y')^2}\ dx \text{ for rotation about the x-axis}\\ S=2\pi\int_0^1 x\sqrt{1+(y')^2}\ dx \text{ for rotation about the y-axis} $$
Here, we notice that for rotation about the y-axis the integration limits are $x\in [0,1]$, since that is the only part of the curve that is being rotated.
Now, it seems that someone has carefully designed your problem to be tractable, because $$\sqrt{1+(y')^2}=\sqrt{1+\sinh^2(4x)}=\cosh(4x)$$
I'll give you my final results (and I hope they are correct!).
$$ S=\frac{\pi}{2}\left[1+\frac{\sinh(8)}{8} \right] \text{ for rotation about the x-axis}\\ S=\frac{\pi}{8}[4\sinh(4)-\cosh(4)+1] \text{ for rotation about the y-axis} $$
Let me know if you need any more detail.
