Suppose we have a vector field $f(x, y)=(-y, x)$ and a curve $C$ that is parameterized by $\alpha(t)=(\cos (t), \sin (t))$ for $0

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

∻Solution by Steps∻ ‖ step 1 ⋮ To find the line integral of the vector field $f(x, y) = (-y, x)$ along the curve $C$ parameterized by $\alpha(t) = (\cos(t), \sin(t))$, we first need to compute $d\alpha$. ‖ ‖ step 2 ⋮ The differential $d\alpha$ is given by the derivative of $\alpha(t)$ with respect to $t$, which is $d\alpha = (-\sin(t), \cos(t))dt$. ‖ ‖ step 3 ⋮ We then compute the dot product $f \cdot d\alpha = (-y, x) \cdot (-\sin(t), \cos(t))dt$. Substituting $x = \cos(t)$ and $y = \sin(t)$, we get $f \cdot d\alpha = (\sin^2(t) + \cos^2(t))dt$. ‖ ‖ step 4 ⋮ Since $\sin^2(t) + \cos^2(t) = 1$, the dot product simplifies to $f \cdot d\alpha = dt$. ‖ ‖ step 5 ⋮ The line integral $\int_{C} f \cdot d\alpha$ is then the integral of $dt$ over the interval $0 < t < 2\pi$. ‖ ‖ step 6 ⋮ Evaluating the integral, we get $\int_{0}^{2\pi} dt = 2\pi - 0 = 2\pi$. ‖ ∻[1] Answer∻ ⚹ $2\pi$ ⚹ ∻Key Concept∻ ⚹Line Integral of a Vector Field⚹ ∻Explanation∻ ⚹The line integral of a vector field along a curve is found by taking the dot product of the vector field and the differential of the curve's parameterization, then integrating over the specified interval.⚹◊What is the formula for calculating the line integral of a vector field along a curve in 2D space?⍭ Generate me a similar question◊

Question

Suppose we have a vector field $f(x, y)=(-y, x)$ and a curve $C$ that is parameterized by $\alpha(t)=(\cos (t), \sin (t))$ for $0<t<2 \pi$.

What is the line integral of $f$ along $C$ ?
$$
\int_{C} f \cdot d \alpha=
$$
$\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the line integral of the vector field $f(x, y) = (-y, x)$ along the curve $C$ parameterized by $\alpha(t) = (\cos(t), \sin(t))$, we first need to compute $d\alpha$. ‖ ‖ step 2 ⋮ The differential $d\alpha$ is given by the derivative of $\alpha(t)$ with respect to $t$, which is $d\alpha = (-\sin(t), \cos(t))dt$. ‖ ‖ step 3 ⋮ We then compute the dot product $f \cdot d\alpha = (-y, x) \cdot (-\sin(t), \cos(t))dt$. Substituting $x = \cos(t)$ and $y = \sin(t)$, we get $f \cdot d\alpha = (\sin^2(t) + \cos^2(t))dt$. ‖ ‖ step 4 ⋮ Since $\sin^2(t) + \cos^2(t) = 1$, the dot product simplifies to $f \cdot d\alpha = dt$. ‖ ‖ step 5 ⋮ The line integral $\int_{C} f \cdot d\alpha$ is then the integral of $dt$ over the interval $0 < t < 2\pi$. ‖ ‖ step 6 ⋮ Evaluating the integral, we get $\int_{0}^{2\pi} dt = 2\pi - 0 = 2\pi$. ‖ ∻[1] Answer∻ ⚹ $2\pi$ ⚹ ∻Key Concept∻ ⚹Line Integral of a Vector Field⚹ ∻Explanation∻ ⚹The line integral of a vector field along a curve is found by taking the dot product of the vector field and the differential of the curve's parameterization, then integrating over the specified interval.⚹◊What is the formula for calculating the line integral of a vector field along a curve in 2D space?⍭ Generate me a similar question◊

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