Calculus 3 Stokes Throrem

Question

Let 

Use Stokes' Theorem to evaluate , where  is the triangle with vertices (5,0,0), (0,5,0), and (0,0,5), oriented counter clockwise as viewed from above.

Answer 1

You are asked to find a line integral around the edge of a triangle. Stokes theorem tells you that that line integral is the same thing as the surface integral of curl(F) through the triangular surface.

First, we must find the curl of the vector field

Let F = <P,Q,R> and P_x denote the partial derivative of P with respect to x.

The formula for curl tells us that curl(F) can be found by

curl(F) = <R_y-Q_z, P_z-R_x, Q_x - P_y>

The partial derivatives give R_y = 0, Q_z = 2z, P_z = 0, R_x = 2x, Q_x = 0, P_y = 2y

curl(F) = <-2z,-2x,-2y>

Next we must find the unit normal vector to the surface

The surface is a plane containing the points (5,0,0);(0,5,0); and (0,0,5). The plane described by these points is x+y+z=5, or z=5-x-y.

Planes are defined by a point and a normal vector where the equation of the plane can be given as

a(x-x_0)+b(y-y_0)+c(z-z_0) = 0, where (x_0,y_0,z_0) is a point on the plane, and <a,b,c> is a vector normal to the plane. From x+y+z = 5, we can see that a normal vector is <1,1,1>

To find the unit normal vector, we divide <1,1,1> by its magnitude (sqrt(1^2+1^2+1^2) = sqrt(3) to get the unit normal vector of (1/sqrt(3))<1,1,1>

Let us begin setting up the surface integral

∫_c (F*dr) = ∫∫_B curl(F)*n dS (where B is the surface bounded by c)

we can find the dot product of curl(F) and n

<-2z,-2x,-2y>*(1/sqrt(3))<1,1,1> = (-2/sqrt(3))(x+y+z)

since we are only concerned with the vector field on the plane, we can set z=5-x-y to get the double integral in terms of 2 variables:

curl(F)*n = (-2/sqrt(3))(x+y+5-x-y) = -10/sqrt(3)

Because this result is a constant, it can be pulled out of the integral

∫∫_B curl(F)*n dS = (-10/sqrt(3))* ∫∫_B dS

The integral is just the surface area of the triangle.

Now, we find the relationship between the dS and dA component to do the integral

dS represents the surface element in 3D space. This is difficult to integrate over by itself. However, there is a direct relationship in Cartesian coordinates between the surface area dS, and dA, the area of the projection of dS into the x-y plane. This is:

dS = sqrt( 1 + (z_x)^2 + (z_y)^2 ) dA (where z_x and z_y can be calculated from the equation of the surface z=5-x-y. This gives

dS = sqrt(1 + (-1)^2 + (-1)^2) dA = sqrt(3)dA

Substituting gives

(-10/sqrt(3))* ∫∫_B dS = (-10/sqrt(3))* ∫∫_D sqrt(3) dA = -10 ∫∫_D dA (where D is the projected region of the triangle into the x-y plane)

this integral is now a double integral over a region in the x-y plane bounded by x=0, y=0, and x+y=5

Finally, we have a double integral with given limits of integration

-10 ∫∫_D dA = -10 ∫_0^5 ∫_0^(5-x) dy dx ( where _a^b denotes the limits of integration)

Evaluating the double integral gives

-10 ∫_0^5 ∫_0^(5-x) dy dx = -10 ∫_0^5 [ y]_0^(5-x) dx

= -10 ∫_0^5 5-x dx = -10 [5x-(x^2)/2]_0^5 = -10[25-25/2 - (0-0)] = -10[25/2] = -125

The final result of the surface integral is -125