Question

Math

Posted 6 months ago

```
Suppose we have a scalar field $f(x, y)=-y$ and a curve $C$ that is parameterized by $\alpha(t)=(\cos (t), \sin (t)-1)$ for $0<t<2 \pi$.
What is the line integral of $f$ along $C$ ?
$\int_{C} f d s=$
$\square$
```

Sign in to unlock the answer

Answer from Sia

Posted 6 months ago

Solution by Steps

step 1

To find the line integral of the scalar field $f(x, y) = -y$ along the curve $C$ parameterized by $\alpha(t) = (\cos(t), \sin(t) - 1)$, we first need to compute the differential arc length $ds$

step 2

The differential arc length is given by $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. We calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ from the parameterization

step 3

For $\alpha(t) = (\cos(t), \sin(t) - 1)$, we have $\frac{dx}{dt} = -\sin(t)$ and $\frac{dy}{dt} = \cos(t)$

step 4

Substituting these into the expression for $ds$, we get $ds = \sqrt{(-\sin(t))^2 + (\cos(t))^2} dt = \sqrt{1} dt = dt$

step 5

The line integral is then $\int_{C} f ds = \int_{0}^{2\pi} -(\sin(t) - 1) dt$ since $f(x, y) = -y$ and $y = \sin(t) - 1$

step 6

Evaluating the integral, we get $\int_{0}^{2\pi} -(\sin(t) - 1) dt = [-\cos(t) - t]_{0}^{2\pi}$

step 7

Plugging in the limits of integration, we find $[-\cos(2\pi) - 2\pi] - [-\cos(0) - 0] = [-1 - 2\pi] - [1 - 0] = -2\pi - 2$

[1] Answer

$-2\pi - 2$

Key Concept

Line Integral of a Scalar Field

Explanation

The line integral of a scalar field along a curve is found by integrating the scalar field function multiplied by the differential arc length over the parameter interval of the curve.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Unlock Smarter Learning with AskSia Super!

Join Super, our all-in-one AI solution that can greatly improve your learning efficiency.

30% higher accuracy than GPT-4o

Entire learning journey support

The most student-friendly features

Study Other Question