evaluated at upper bound 1 and lower bound 0
evaluated at upper bound 1 and lower bound 0
evaluated at upper bound 1 and lower bound 0
evaluated at upper bound 1 and lower bound 0
- Is this right?
evaluated at upper bound 1 and lower bound 0
evaluated at upper bound 1 and lower bound 0
evaluated at upper bound 1 and lower bound 0
evaluated at upper bound 1 and lower bound 0
- Is this right?
Why have you written your derivative in such an odd format?
$\displaystyle \begin{align*} y &= \sqrt{ 2 - x^2} \\ \frac{ \mathrm{d}y }{\mathrm{d}x} &= -\frac{2x}{2 \sqrt{2 - x^2} } \\ &= -\frac{x}{\sqrt{2 - x^2}} \end{align*}$
so your arclength is
$\displaystyle \begin{align*} L &= \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \, \mathrm{d}x } \\ &= \int_0^1{ \sqrt{ 1 + \left( -\frac{x}{\sqrt{2 - x^2} } \right) ^2 } \, \mathrm{d}x } \end{align*}$
Go from here...
Fixed that step
NO! Here your $\displaystyle \begin{align*} a = \sqrt{2} \end{align*}$ so the substitution you must make is $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \implies \mathrm{d}x = \sqrt{2}\cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$
Using trig substitution you get to this point:
Even though you are out by a significant factor from the above mistake, you come to the right conclusion that the "functional" part cancels out. So how would you evaluate $\displaystyle \begin{align*} \int{ \mathrm{d}x } \end{align*}$? Hint: $\displaystyle \begin{align*} \int{ \mathrm{d}x} = \int{ 1 \, \mathrm{d}x } \end{align*}$...
- What next it seems like all the thetas cancel out, so how can we evaluate it?
Well, if you mean evaluated where x is between 0 and 1, then yes.
But you will be better off changes your bounds to $\displaystyle \begin{align*} \theta \end{align*}$ bounds, so that you don't have to convert back to a function of x.
You know that $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}$. If x = 0 then what is $\displaystyle \begin{align*} \theta \end{align*}$? If x = 1 then what is $\displaystyle \begin{align*} \theta \end{align*}$?
NO! You are trying to figure out which ANGLES give those sine values! What angle gives the sine value as 0? (You have correctly identified it as 0). What angle gives $\displaystyle \begin{align*} \sqrt{2}\sin{ \left( \theta \right) } = 1 \implies \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}$. This is another one that you should have committed to memory.