Answer
2
Work Step by Step
x=\int_{y}^{0}\sqrt{sec^4(t)-1}dt on -\frac{\pi}{4}\leq y\leq \frac{\pi}{4}
Consider: x=f(y)
In order to find the length of the curve, use the arclength formula:
If x=f(y) is continuous and differentiable on c\leq y \leq d
then {\color{Red} L=\int_{c}^{d}\sqrt{1+(\frac{dx}{dy})^2}dy }
1) Find \frac{dx}{dy}
\frac{\mathrm{d}}{\mathrm{d}y} \int_{0}^{y} \sqrt{sec^4(t)-1}dt=\sqrt{sec^4(y)-1}
because of Part 1 of the Fundamental Theorem of Calculus: \frac{\mathrm{d}}{\mathrm{d}x} \int_{a}^{x} f(t)dt=f(x)
2) Plug \frac{dx}{dy} into the arclength formula
L=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{1+(\sqrt{sec^4(y)-1})^2}dy
3) Solve
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{1+sec^4(y)-1}dy=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{sec^4(y)}dy=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}sec^2(y)dy=[tan(y)]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=tan(\frac{\pi}{4})-tan(-\frac{\pi}{4})=1-(-1)=1+1=2
