Volume $y=\cosh x$ rotated around $y$-axis

Question

How do I find the volume of $y=\cosh x$ rotating around the $y$-axis from 0 to 1.

I know the washer method involves solving for $x$. But in this question I cannot solve for $x$.

$\cosh x = (e^x+e^{-x})/2$. How can I write this in terms of $x$?

Answer 1

You want to calculate the $$\int_{x=0}^{x=1}{2πx\cosh x\mathrm d x}.$$

What can you now do with integrating by parts?

Answer 2

The correct shell setup is

$$2\pi\int_0^1 x[cosh(1)-cosh(x)]dx$$

which leads to the result

$$\frac{\pi}{2}(e+5e^{-1}-4)$$

Answer 3

You asked originally about solving $cosh(x)= \frac{e^{x}+ e^{-x}}{2}= y$ for x. I hope you know that $a^{-b}= \frac{1}{a^b}$ so that $e^{-x}= \frac{1}{e^x}$. Using that we can write the equation as $\frac{e^{x}+ \frac{1}{e^x}}{2}= y$.

An obvious first step is to multiply both sides by 2: $e^{x}+ \frac{1}{e^x}= 2y$. Get rid of the fraction by multiplying both sides by $e^x$ to get $(e^x)^2+ 1= 2e^xy$ or $(e^x)^2- 2y e^x+ 1= 0$. That's a quadratic equation in $e^x$ and we can solve it using the quadratic formula: $e^x= \frac{2\pm\sqrt{4y^2- 4}}{2}= 1\pm\sqrt{y^2- 1}$. Finally, $x= ln\left(1\pm\sqrt{y^2- 1}\right)$.

The problem is that "cosh(x)" itself is not "one-to-one" so does not have a true inverse. You have to decide which branch to use- whether to use the "+" or "-". Here, since you want the part that is in the first quadrant, take the "+": $x= ln\left(1+ \sqrt{y^2- 1}\right)$.