QuestionEvaluate $\oint(y z d x+x z d y+x y d z)$ along the helix $x=a \cos t, y=a \sin t, z=k t$ for $t$ from 0 to $2 \pi$ .

Studdy Solution

STEP 1

Assumptions1. The given integral is a line integral over the arc of a helix. . The parametric equations for the helix are $x=a \cos t$ , $y=a \sin t$ , and $z=k t$ .
3. The parameter $t$ varies from0 to $\pi$ .

STEP 2

We need to compute the differential elements $dx$ , $dy$ , and $dz$ in terms of $dt$ . We can do this by differentiating the parametric equations for $x$ , $y$ , and $z$ with respect to $t$ .
$dx = \frac{dx}{dt} dt = -a \sin t dt$ $dy = \frac{dy}{dt} dt = a \cos t dt$ $dz = \frac{dz}{dt} dt = k dt$

STEP 3

Substitute the parametric equations for $x$ , $y$ , and $z$ , and the differential elements $dx$ , $dy$ , and $dz$ into the integral.
$\oint(y z dx + x z dy + x y dz) = \oint(a \sin t \cdot k t \cdot -a \sin t dt + a \cos t \cdot k t \cdot a \cos t dt + a \cos t \cdot a \sin t \cdot k dt)$

STEP 4

implify the integral.
$\oint(y z dx + x z dy + x y dz) = -a^2 k \oint t \sin^2 t dt + a^2 k \oint t \cos^2 t dt + a^2 k \oint \cos t \sin t dt$

STEP 5

Now, we need to evaluate these integrals over the interval $t$ varies from0 to $2 \pi$ .

STEP 6

The first two integrals can be evaluated using the power-reduction identities $\sin^2 t = \frac{1 - \cos2t}{2}$ and $\cos^2 t = \frac{1 + \cos2t}{2}$ .

STEP 7

Substitute these identities into the first two integrals and simplify.
$-a^2 k \oint t \sin^2 t dt = -a^2 k \oint t \frac{1 - \cos2t}{2} dt = -\frac{a^2 k}{2} \oint t dt + \frac{a^2 k}{2} \oint t \cos2t dt$ $a^2 k \oint t \cos^2 t dt = a^2 k \oint t \frac{1 + \cos2t}{2} dt = \frac{a^2 k}{2} \oint t dt + \frac{a^2 k}{2} \oint t \cos2t dt$

STEP 8

The integral $\oint t dt$ over the interval $t$ varies from0 to $2 \pi$ is simply $\frac{1}{2} (2 \pi)^2 =2 \pi^2$ .

STEP 9

The integral $\oint t \cos2t dt$ over the interval $t$ varies from to $2 \pi$ is, because the cosine function has a period of $2 \pi$ and its integral over one period is.

STEP 10

Substitute these results into the first two integrals.
$-\frac{a^2 k}{2} \oint t dt + \frac{a^2 k}{2} \oint t \cos2t dt = -\frac{a^2 k}{2} \cdot2 \pi^2 + \frac{a^2 k}{2} \cdot0 = -a^2 k \pi^2$ $\frac{a^2 k}{2} \oint t dt + \frac{a^2 k}{2} \oint t \cos2t dt = \frac{a^2 k}{2} \cdot2 \pi^2 + \frac{a^2 k}{2} \cdot0 = a^2 k \pi^2$

STEP 11

The third integral can be evaluated using the double-angle identity \cos t \sin t = \frac{}{} \sint.

STEP 12

Substitute this identity into the third integral and simplify.
$a^2 k \oint \cos t \sin t dt = \frac{a^2 k}{2} \oint \sin2t dt$

STEP 13

The integral $\oint \sin2t dt$ over the interval $t$ varies from0 to $2 \pi$ is0, because the sine function has a period of $2 \pi$ and its integral over one period is0.

STEP 14

Substitute this result into the third integral.
$\frac{a^2 k}{2} \oint \sin2t dt = \frac{a^2 k}{2} \cdot0 =0$

STEP 15

Finally, add up the results of the three integrals to get the value of the original integral.
$\oint(y z dx + x z dy + x y dz) = -a^2 k \pi^2 + a^2 k \pi^2 +0 =0$ So, the value of the integral over the arc of the helix is0.

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