Question

evaluate the line integral of f(x,y) along the curve C

Answer 1

f(x,y)=x^2\[\div \sqrt{1+4y},C:y=x^2,0\le x \le3\]

Answer 2

\[\int\limits_{0}^{9}\sqrt{1+f'(y)^{2}} dy\]
where \[f(y) = \frac{y}{\sqrt{1+4y}}\]

does that look right

Answer 3

yeah it looks right

Answer 4

but im really not sure

Answer 5

ok
\[f'(y) = \frac{2y+1}{(1+4y)\sqrt{1+4y}}\]
\[f'(y)^{2} = \frac{(2y+1)^{2}}{(1+4y)^{3}}\]
substitute that into integral
hold on i haven't solved it further yet

Answer 6

hmm it can't be integrated so i probably did something wrong

Answer 7

ok i know what i did wrong...i did the integral for finding arc length which is different than the line integral.
you are supposed to parametrize f(x,y) using parametric equations
Let
x = t
y = t^2
\[f(t) = \frac{t^{2}}{\sqrt{1+4t^{2}}}\]
the line integral is
\[\int\limits_{a}^{b}f(r(t))|r'(t)| dt\]
r(t) = t^2
r'(t) = 2t
\[\rightarrow \int\limits_{0}^{3} \frac{t^{2}}{\sqrt{1+4t^{2}}}(2t) dt\]
To integrate use substitution
\[u = 1+4t^{2}\]
\[du = 8t\]
\[\rightarrow \frac{1}{16}\int\limits_{0}^{3}\frac{u-1}{\sqrt{u}}du\]\[= \frac{1}{16}\int\limits_{0}^{3}\sqrt{u} - \frac{1}{\sqrt{u}} du\]
\[=\frac{1}{16}((2/3)u^{3/2} - 2u^{1/2})\]
\[=\frac{1}{16}((2/3)(1+4t^{2})^{3/2} - 2(1+4t^{2})^{1/2}) from 0 \to 3\]
\[\approx 8.7\]