Question

Evaluate the following integral using Stokes theorem

Integralof over C (x+z)dx+(x-y)dy+xdz)

Where C is an ellipse

(x^2/16)+(y^2/25)=1 z=4

Answer 1

We have $I=\oint\limits_C(x+z)dx+(x-y)dy+xdz$.

The Stokes' formula is:

$\oint\limits_C(Pdx + Qdy + Rdz)=\iint\limits_S(\frac{\partial{R}}{\partial{y}}-\frac{\partial{Q}}{\partial{z}})dydz+$

$+(\frac{\partial{P}}{\partial{z}}-\frac{\partial{R}}{\partial{x}})dzdx + (\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}})dxdy$.

Where $S -$ the part of the plane $(z=4)$ bounded by an ellipse $C$.

And $C$ is the ellipse $(\frac{x^2}{16}+\frac{y^2}{25} = \frac{x^2}{4^2}+\frac{y^2}{5^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2})$.

And $P = x+z$, $Q=x-y$, $R=x$.

Find the necessary partial derivatives:

$\frac{\partial{R}}{\partial{y}}=0, \frac{\partial{Q}}{\partial{z}}=0, \frac{\partial{P}}{\partial{z}}=1, \frac{\partial{R}}{\partial{x}}=1, \frac{\partial{Q}}{\partial{x}}=1, \frac{\partial{P}}{\partial{y}}=0.$

So, we can write:

$\iint\limits_S(0-0)dydz+\iint\limits_S(1-1)dzdx+\iint\limits_S(1-0)dxdy$.

$\iint\limits_Sdxdy = I$.

The integrand is equal to one $(f(x, y)=1)$, so the double integral is equal to the surface area $S$.

It means that the $I$ is equal to the area of the ellipse:

$I=\pi ab=4\times5\times\pi=20\pi.$

Answer: $I=20\pi \approx62.8$