calculusmultivariable-calculus
Find \begin{align*} \int_C \sqrt{1+4x^2 z^2} ds, \end{align*} where $C$ is the curve of intersection of the surfaces $x^2 + z^2 = 1$ and $y = x^2$.
Attempt at solution: So first I need a parametrization of this curve. I let $x = t$. Then we have $y = t^2$ and $z = +- \sqrt{1-t^2}$. But I'm not sure what sign I should pick here, and what my integration bounds are?
Any help would be appreciated.
As you've found now, the arclength integration on the quarter-circle (using what is essentially angle-parametrization) in the second quadrant is
$$ \int_{\pi / 2}^{\pi} \ \ (2 \ \sin \ t)^2 \ \ \sqrt{(-2 \sin \ t)^2 \ + \ (2 \cos \ t)^2} \ \ dt \ \ = \ \ \int_{\pi / 2}^{\pi} \ \ 4 \ \sin^2 t \ \cdot \ 2 \ \ dt \ \ $$
$$ = \ \ 8 \ \int_{\pi / 2}^{\pi} \ \ \frac{1}{2} ( \ 1 \ - \ \cos \ 2t \ )\ \ dt \ \ = \ \ ( \ 4 \ t \ - \ 2 \ \sin \ 2 t \ ) \ \vert_{\pi / 2}^{\pi} $$
$$ = \ \ ( \ 4 \ \pi \ - \ 2 \ \sin \ 2 \ \pi \ ) \ - \ ( \ 4 \ \cdot \frac{\pi}{2} \ - \ 2 \ \sin \ 2 \cdot \frac{\pi}{2} \ ) $$
$$ = \ 4 \ \pi \ - \ 0 \ - \ 2 \ \pi \ + \ 0 \ = \ 2 \ \pi \ \ . $$
Here is a graph of the tangent vector $ \ \mathbf{r}(t) \ $ following the circular arc. Both $ \ dx \ $ and $ \ dy \ $ are negative in the second quadrant $ ^* $, since the $ \ x \ $ and $ \ y \ $ coordinates of points along the circle are decreasing as the vector "moves" in the specified direction. Nonetheless, $ \ ds \ $ , the infinitesimal arclength element is always positive; since $ \ y^2 \ $ is non-negative, the result of the integration should have a positive value.
$ ^* $ The fact that $ \ dy \ $ is negative on this arc is the reason ellya obtained a negative value from integrating $ \ \int_C \ y^2 $ dy .
The reason you only obtained the plane after step 1 is because you have only carried out one step in the solution of the system of 2 equations. Since your system has more variables than equations, you will need to proceed further using the results from your first step.
For example, substituting the plane $y=\frac{2-6x}{5}$ inside either of the equations gives an expression involving $z$ and $y$.
Best Answer
Parameterize a curve by letting $x=\cos u$, $z = \sin u$, and $y=\cos^2 u$, where $-\pi \le u \le \pi$.
Then,
$$\begin{align} ds &=\sqrt{x'(u)^2+y'(u)^2+z'(u)^2}\, du\\\\ &=\sqrt{\sin^2u+4\cos^2u\sin^2u+\cos^2u}\, du\\\\ &=\sqrt{1+4\cos^2u\sin^2u}\, du \end{align}$$
The integral becomes
$$\int_{-\pi}^{\pi} \left(1+4\cos^2 u \sin^2 u\right) du$$