Find the length of the piece of this curve where $x \geq \frac{3}{2}$

Question

Consider the curve C which is the intersection of the two cylinders of equations $e^z=x$ and $x^2+y^2=2x$. Find the length of the piece of this curve where $x \geq \frac{3}{2}$

I have done the parameterization, but when I apply the Length formula I am getting something that is difficult to integrate. Also, I know that the portion of curve I am measuring is when $x \geq \frac{3}{2}$, but how do I make that in terms of 't' to get the range for my interval? I had thought to substitute $x = \frac{3}{2}$ for my $x$ term in the parameterization, and that gives me $t=\pi/3$ and $5/pi/3$ when $x=\frac{3}{2}$ Does that seem correct and is my parameterization correct?

Answer 1

Looks ok to me except for a bit strange argument with $x=3/2$ and $t=\pi/3$ and $5\pi/3$. You need to solve $x\ge 3/2$, not $=3/2$ and find the corresponding integral bounds for $t$ (btw they are missing on the picture!). Another thing: you still have to find an anti-derivative for the integrand. How do you think to proceed? Hint: find a good trigonometric formula to get rid of the root sign.

Answer 2

Most of your paper is correct, at the end you would have

$$ \int_{-\pi/3}^{\pi/3} \sqrt{\frac{2}{\cos(t)+1}} dt. $$

Now use

$$ \cos(2x) = \cos^2(x) - \sin^2(x) = 2 \cos^2(x) - 1, $$

so

$$ \cos(2x) + 1 = 2 \cos^2(x), $$

so you can write

$$ \int_{-\pi/3}^{\pi/3} \frac{1}{\cos(t/2)} dt, $$

which is easier to solve...

$$ \int_{-\pi/3}^{\pi/3} \frac{1}{\cos(t/2)} dt = \int_{-\pi/3}^{\pi/3} \frac{\cos(t/2)}{\cos^2(t/2)} dt = 2 \int_{-\pi/3}^{\pi/3} \frac{1}{1 - \sin^2(t/2)} d\sin(t/2) \\= \int_{-\pi/3}^{\pi/3} \left( \frac{1}{1 - \sin(t/2)} + \frac{1}{1 + \sin(t/2)} \right) d\sin(t/2) \\= \Big[ \ln( 1 + \sin(t/2) ) - \ln( 1 - \sin(t/2) ) \Big]_{-\pi/3}^{\pi/3} \\= 2 \ln(3) $$