Question

Use Stokes' Theorem to evaluate int_C F*dr where C is oriented counterclockwise as viewed above.F(x,y,z)=xy i + 3zj+5yk, C is the curve of intersection of the plane x+z=10 and the cylinder x^2+y^2=9.

Use Stokes' Theorem to evaluate \(\int_C F \cdot dr\) where C is oriented counterclockwise as viewed above.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}\), C is the curve of intersection of the plane \(x+z=10\) and the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={9}\).

Answers (1)

2021-01-07

Step 1
Here use the stokes theorem to find the integral of the function.
And given that the function \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}\)
The curve of intersection of the plane \(x + z =10\) and the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={9}\)
Step 2
The Stokes theorem is given as:
\(\displaystyle\int_{{C}}{F}\cdot{d}{r}=\int\int_{{S}}{c}{u}{r}{l}{F}\cdot{d}{s}\)
if \(\displaystyle{F}={P}\hat{{{i}}}+{Q}\hat{{{j}}}+{R}\hat{{{k}}}\)
Then \(\displaystyle{C}{u}{r}{l}{F}={\left({R}_{{y}}-{Q}_{{z}}\right)}\hat{{{i}}}+{\left({P}_{{z}}-{R}_{{x}}\right)}\hat{{{j}}}+{\left({Q}_{{x}}-{P}_{{y}}\right)}\hat{{{k}}}\)
Given \(\displaystyle{F}={x}{y}\hat{{{i}}}+{3}{z}\hat{{{j}}}+{5}{y}\hat{{{k}}}\)
Therefore,
Curl \(\displaystyle{F}={\left({5}-{3}\right)}\hat{{{i}}}+{\left({0}-{0}\right)}\hat{{{j}}}+{\left({0}-{x}\right)}\hat{{{k}}}={2}\hat{{{i}}}-{x}\hat{{{k}}}\)
Step 3
Let S be the path of the plane \(x + z =10\) inside the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={9}\)
If the surface S is of the form \(z = g(x, y)\) and is oriented upwards,
So, by the formula:
\(\displaystyle\int\int_{{S}}{F}\cdot{d}{S}=\int\int_{{D}}-{P}{\left(\frac{{\partial{g}}}{{\partial{x}}}-{Q}\frac{{\partial{g}}}{{\partial{y}}}\right)}+{R}{d}{A}\)
Here S is the part of the plane \(z = 10-x\)
And D is region inside the circle \(\displaystyle{x}^{{2}}+{y}^{{2}}={0},{z}={0}\)
\(\displaystyle\int\int_{{S}}{C}{u}{r}{l}{F}\cdot{d}{s}=\int\int_{{D}}{\left({2}-{x}\right)}{d}{A}\)
Step 4
Solve further:
\(\displaystyle\int\int_{{S}}{F}\cdot{d}{S}=={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{3}}}{\left({2}-{r}^{{2}}{\cos{{0}}}\right)}{r}{d}{r}{d}{0}\)
\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\left[{\int_{{0}}^{{3}}}{2}{r}{d}{r}-{\int_{{0}}^{{3}}}{r}^{{3}}{\cos{{\left({0}\right)}}}{d}{r}\right]}{d}{0}\)
\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{r}^{{2}}}{{2}}+\frac{{r}^{{4}}}{{4}}{\cos{{0}}}\right]}_{{0}}^{{3}}}{d}{0}\)
\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\left[{9}-{\cos{{\left({0}\right)}}}\frac{{81}}{{4}}\right]}{d}{0}\)
\(\displaystyle={18}\pi-{0}={56.54}\)

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