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Lapatulllka [165]
3 years ago
15

Find the counterclockwise circulation and outward flux of the field F=7xyi+5y^2j around and over the boundary of the region C en

closed by the curves y=x^2 and y=x in the first quadrant.
Mathematics
1 answer:
dezoksy [38]3 years ago
5 0

Split up the boundary of <em>C</em> (which I denote ∂<em>C</em> throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to <em>y</em> = <em>x</em> ²), and the line segment from (1, 1) to (0, 0) (the part of ∂<em>C</em> on the line <em>y</em> = <em>x</em>).

Parameterize these pieces respectively by

<em>r</em><em>(t)</em> = <em>x(t)</em> <em>i</em> + <em>y(t)</em> <em>j</em> = <em>t</em> <em>i</em> + <em>t</em> ² <em>j</em>

and

<em>s</em><em>(t)</em> = <em>x(t)</em> <em>i</em> + <em>y(t)</em> <em>j</em> = (1 - <em>t</em> ) <em>i</em> + (1 - <em>t</em> ) <em>j</em>

both with 0 ≤ <em>t</em> ≤ 1.

The circulation of <em>F</em> around ∂<em>C</em> is given by the line integral with respect to arc length,

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds

where <em>T</em> denotes the <em>tangent</em> vector to ∂<em>C</em>. Split up the integral over each piece of ∂<em>C</em> :

• on the parabolic segment, we have

<em>T</em> = d<em>r</em>/d<em>t</em> = <em>i</em> + 2<em>t</em> <em>j</em>

• on the line segment,

<em>T</em> = d<em>s</em>/d<em>t</em> = -<em>i</em> - <em>j</em>

Then the circulation is

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}

Alternatively, we can use Green's theorem to compute the circulation, as

\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}

The flux of <em>F</em> across ∂<em>C</em> is

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds

where <em>N</em> is the <em>normal</em> vector to ∂<em>C</em>. While <em>T</em> = <em>x'(t)</em> <em>i</em> + <em>y'(t)</em> <em>j</em>, the normal vector is <em>N</em> = <em>y'(t)</em> <em>i</em> - <em>x'(t)</em> <em>j</em>.

• on the parabolic segment,

<em>N</em> = 2<em>t</em> <em>i</em> - <em>j</em>

• on the line segment,

<em>N</em> = - <em>i</em> + <em>j</em>

So the flux is

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}

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