Question

Integrate f over the given curve. f(x,y) = x³/y, C: y= x²/2, 0≤x≤1

Answer 1

The line integral of
`\(f(x, y) = (x^3)/(y)\)` over the curve
`\(C: y = (x^2)/(2)\)`,
`\(0 \leq x \leq 1\)`, is
`\((1)/(12)\)` units.

Step-by-step explanation:

To compute the line integral of
`\(f(x, y)\)` over the given curve
`\(C\)`, we need to parameterize the curve in terms of a single variable, usually denoted as
`\(t\)`. In this case, since
`\(C\)` is described by
`\(y = (x^2)/(2)\)` and
`\(0 \leq x \leq 1\)`, we can rewrite this equation in parametric form as
`\(x = t\) and \(y = (t^2)/(2)\)`, where
`\(t\)` ranges from
`\(0\) to \(1\)`.

Next, we need to express
`\(f(x, y)\)` in terms of
`\(t\)` using the parameterization. Substituting
`\(x = t\)` and
`\(y = (t^2)/(2)\)` into
`\(f(x, y) = (x^3)/(y)\)`, we get
`\(f(t) = (t^3)/((t^2)/(2)) = 2t\)`.

The line integral of
`\(f(x, y)\)` over
`\(C\)` is then given by the integral of
`\(f(t)\)` with respect to
`\(t\)` from
`\(0\) to \(1\)`:

`\(\int_(C) f(x, y) \,ds = \int_(0)^(1) f(t) \cdot \| \mathbf{r}'(t) \| \,dt\)`, where
`\(\mathbf{r}(t) = \langle t, (t^2)/(2) \rangle\)` is the parametric representation of
`\(C\)`, and
`\(\| \mathbf{r}'(t) \|\)` represents the magnitude of the derivative of
`\(\mathbf{r}(t)\)`.

The derivative of
`\(\mathbf{r}(t)\) is \(\mathbf{r}'(t) = \langle 1, t \rangle\)`, and its magnitude is
`\(\| \mathbf{r}'(t) \| = √(1 + t^2)\)`.

Evaluating the integral
`\(\int_(0)^(1) f(t) \cdot \| \mathbf{r}'(t) \| \,dt = \int_(0)^(1) 2t \cdot √(1 + t^2) \,dt\)` yields the result of
`\((1)/(12)\)` units.