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Let $\gamma_1$ be the intersection curve between the surface $x^2+(y+z)+z^2=1$ and the plane $z=x$ and let $\gamma_2$ be $\gamma_1$ with $x\geq0$,$y\geq0$. Calculate $$\int_{\gamma}(z^3+2yz^5)dx+(x^3+z^6)dy+4yz^5dz$$

I need someone to see if my argument about the curves are correct. I have only post the part there I write about how to parametrize the intersection curve and how to get the curve to get closed.

A part of my solution

To use the Stokes's theorem I need to have a bounded curve, but because of the conditions $x \geq 0$,$y \geq 0$ the curve is not closed. Therefore I need to add two curves such that I get and closed curve to the surface $Y$ which has the bounded curves as its boundary so that the surface is also bounded/closed ?

To get the 2 curves I need I can first let $y=0$ and then $z=0=x$ right?

This gives me the curves

$$\gamma_3 : y^2 \leq 1, \; z=0=x, \; y \geq 0 $$

$$\gamma_4 : 3x^2 \leq 1, \; z=0=x, \; x \geq 0 $$ which is equivalent with the lines $$\gamma_3 : 0\leq y \leq 1, \; z=0=x$$

$$\gamma_4 : 0\leq x \leq \frac{1}{\sqrt{3}}, \; z=x $$ and theirs parametrization are $$\gamma_3 : (x(t),y(t),z(t))=(0,t,0), \quad t \in [0,1] $$ $$\gamma_4 : (x(t),y(t),z(t))=(t,0,t), \quad t \in \left[ 0,\frac{1}{\sqrt{3}} \right] $$ and we have the intersection curve (ellipse) $$\gamma_2 : x^2+(y+x)^2+z^3=1,\; z=x, \; x \geq 0, \; y \geq 0 $$ We can parametrize the surface $Y$ with its boundary $\gamma=\gamma_2+\gamma_3-\gamma_3$ ($\gamma_3$ is negative orientated) like $$(x(u,v),y(u,v),z(u,v))=(u,v,u)$$ which satisfies $$2u^2+(v+u)^2 \leq 1, \; u \geq 0, \; v \geq 0$$ Finally we use Stokes's theorem and get that $$\int_{\gamma_2}{\mathbf F} d {\mathbf r}=\iint_{Y}\operatorname{rot}({\mathbf F}) \cdot {\mathbf N} dS+\int_{\gamma_3}{\mathbf F} d {\mathbf r}-\int_{\gamma_4}{\mathbf F}d {\mathbf r}$$ with the vectorfield $$F=(z^3+2yz^5,x^3+z^6,4yz^5)$$

It is my first time using math stacks, so my solution can be a bit complicated

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  • $\begingroup$ Welcome to MSE, Mona! I've fixed a bit of your LaTeX, I hope you don't mind. I've also removed a tag. $\endgroup$ – Mark Fantini May 30 '14 at 19:48
  • $\begingroup$ thanks. i appreciate it :) @Fantini $\endgroup$ – Mona May 30 '14 at 19:49

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