Using Stokes' theorem, to compute line integrals for the simple close curve from the intersection of Plane and Paraboloid

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How can I apply all these above definitions of line integrals, surface integrals, Stokes theorem to answer the question in #1?
 
I computed the line integral as follows:

Curve[imath] D =(x-1)^2 + (y-1)^2 =1 [/imath]. Now I parametrize it [imath]x= \cos(t) , y=\sin(t) , z = 2\cos(t) + 2 \sin(t) -1[/imath]

[math]F (x,y,z) = zdx + xdy +y dz, r(t) = (\cos(t))i + (\sin(t))j +(\cos(t) +\sin(t))k , x'(t) = - \sin(t), y'(t)= \cos(t), z'(t)= -2\sin(t) +2\cos(t)[/math]
[math] \displaystyle\int_ D F\cdot dr = \displaystyle\int_0^{2\pi} (2\cos(t)+ 2\sin(t)-1)(-\sin(t)) +(\cos(t))(\cos(t))+(\sin(t))(-2\sin(t) + 2\cos(t)) dt [/math][math] \displaystyle\int_ D F\cdot dr = \displaystyle\int_0^{2\pi} \cos^2(t) - 4\sin^2(t) - 4\sin(t)\cos(t) - \sin(t) dt [/math][math] \displaystyle\int_ D F\cdot dr = -3\pi [/math]
 
I will get you started on the surface integral over the paraboloid. The equation of the paraboloid is [imath]z=x^2+y^2[/imath]. Writing it in parametric form using [imath]x[/imath] and [imath]y[/imath] as the parameters is [imath]\vec R(x,y) = \langle x, y, x^2+y^2 \rangle[/imath]. If you calculate it, you will find that [imath]\vec R_x \times \vec R_y = \langle -2x, -2y, 1 \rangle[/imath]. Notice that the [imath]z[/imath] component is positive, pointing upwards, so if you use this normal you will want to traverse the curve counterclockwise as viewed from above. Otherwise use the negative of this vector for [imath]\vec N[/imath]. Using the parametric form you will have:[math]\iint_S \nabla \vec F \cdot \vec N d\sigma = \iint_{(x,y)} \nabla \vec F \cdot \vec R_x\times \vec R_y~dydx =\iint_{(x,y)}\langle 1,1,1\rangle \cdot \langle -2x,-2y,1\rangle dydx[/math] where now you are integrating over the area of the projection of the curve in the [imath]x,y[/imath] plane. To get the boundary of that area, set the[imath]z[/imath] values of the plane and paraboloid equal. You should get [imath]x^2+y^2 =2x+2y-1[/imath], which if you complete the square can be written as [imath](x-1)^2+(y-1)^2 =1[/imath]. This suggests you change to polar coordinates centered at [imath](1,1)[/imath] of [math]x=1+r\cos\theta,~y=1+r\sin\theta,~dydx =rdrd\theta[/math]. Substitute that in, work out the integral, keep you fingers crossed, and I think you will get [imath]3\pi[/imath]. For the plane surface, same idea but easier. Let us know how it comes out for you.
 
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LCKurtz said:
I will get you started on the surface integral over the paraboloid. The equation of the paraboloid is [imath]z=x^2+y^2[/imath]. Writing it in parametric form using [imath]x[/imath] and [imath]y[/imath] as the parameters is [imath]\vec R(x,y) = \langle x, y, x^2+y^2 \rangle[/imath]. If you calculate it, you will find that [imath]\vec R_x \times \vec R_y = \langle -2x, -2y, 1 \rangle[/imath]. Notice that the [imath]z[/imath] component is positive, pointing upwards, so if you use this normal you will want to traverse the curve counterclockwise as viewed from above. Otherwise use the negative of this vector for [imath]\vec N[/imath]. Using the parametric form you will have:[math]\iint_S \nabla \vec F \cdot \vec N d\sigma = \iint_{(x,y)} \nabla \vec F \cdot \vec R_x\times \vec R_y~dydx =\iint_{(x,y)}\langle 1,1,1\rangle \cdot \langle -2x,-2y,1\rangle dydx[/math] where now you are integrating over the area of the projection of the curve in the [imath]x,y[/imath] plane. To get the boundary of that area, set the[imath]z[/imath] values of the plane and paraboloid equal. You should get [imath]x^2+y^2 =2x+2y-1[/imath], which if you complete the square can be written as [imath](x-1)^2+(y-1)^2 =1[/imath]. This suggests you change to polar coordinates centered at [imath](1,1)[/imath] of [math]x=1+r\cos\theta,~y=1+r\sin\theta,~dydx =rdrd\theta[/math]. Substitute that in, work out the integral, keep you fingers crossed, and I think you will get [imath]3\pi[/imath]. For the plane surface, same idea but easier. Let us know how it comes out for you.
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Hi,
As per your method, I solved your integral and I got the answer [imath]-3\pi[/imath] and not [imath]3\pi[/imath].
[math]\displaystyle\int_C F \cdot dr = \displaystyle\int_0^{2\pi} \displaystyle\int_0^1 (-3 -2\cos(\theta)-2\sin(\theta))\cdot rdrd\theta=-3\pi[/math]
Author provided the answer [imath]-3\pi[/imath]. So, your method of answering this question is also correct.
 
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