Line integral

• Apr 26th 2009, 08:28 AM
Apprentice123
Line integral

Calculate the line integral along the curve C

1) $\int_C (x+2y)dx + (x-y)dy$
$C: x = 2cost; y=4sent; (0 \leq t \leq \frac{ \pi}{4})$

My solution

$\int_0^{\frac{ \pi}{4}} (2cost,8sent).(2cost,-4sent)dt$
$\int_0^{\frac{ \pi}{4}} (4cos^2t-32sin^2t)dt$
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.
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$= \frac{-7 \pi}{4} + \frac{9 \sqrt{2}}{8}$

2) $\int_C -ydx + xdy$
$C: y^2 = 3x$
$(3,3)$ to $(0,0)$

My solution

$\int_0^3 (- \sqrt{3t},t).(1,\frac{3}{2 \sqrt{3t}})dt$
.
.
.
$= -3$
• Apr 26th 2009, 10:01 AM
Opalg
Quote:

Originally Posted by Apprentice123

Calculate the line integral along the curve C

1) $\int_C (x+2y)dx + (x-y)dy$
$C: x = 2cost; y=4sent; (0 \leq t \leq \frac{ \pi}{4})$

My solution

$\int_0^{\frac{ \pi}{4}} (2cost,8sent).(2cost,-4sent)dt$
$\int_0^{\frac{ \pi}{4}} (4cos^2t-32sin^2t)dt$
.
.
.
$= \frac{-7 \pi}{4} + \frac{9 \sqrt{2}}{8}$

This doesn't look right at all to me. If $x = 2\cos t$ and $y = 4\sin t$ then $dx = -2\sin t\,dt$ and $dy = 4\cos t\,dt$. The integral is
. . . . . . \begin{aligned}\int_0^{\pi/4}\bigl((2\cos t&{} + 8\sin t)(-2\sin t) + (2\cos t - 4\sin t)(4\cos t)\bigr)\,dt \\ &= \int_0^{\pi/4}(8\cos^2 t - 20\cos t\sin t - 16 \sin^2 t)\,dt \\ &= \ldots = 1-\pi.\end{aligned}

Quote:

Originally Posted by Apprentice123
2) $\int_C -ydx + xdy$
$C: y^2 = 3x$
$(3,3)$ to $(0,0)$

My solution

$\int_0^3 (- \sqrt{3t},t).(1,\frac{3}{2 \sqrt{3t}})dt$
.
.
.
$= -3$

That looks correct, except that if the path goes from (3,3) to (0,0) then the integral should be $\int_3^0\!\!\!\ldots$ rather than $\int_0^3\!\!\!\ldots$ , giving the answer +3 instead of –3.
• Apr 26th 2009, 10:06 AM
Apprentice123
Thank you.
1) I forgot to make the derivative.

2) How do I know the limits of integral having two points?