Let S be the surface of cone z=√x^2 y^2 bounded by the planes z=0 and ...
The Surface S and the Closed Curve C
The surface S is a cone that is bounded by the planes z=0 and z=3. The equation of the cone is z=√(x^2 + y^2). This means that for any point (x, y, z) on the cone, the z-coordinate is equal to the square root of the sum of the squares of the x and y coordinates.
The closed curve C forms the boundary of the surface S. It is the intersection of the cone with the planes z=0 and z=3. The curve C is a circle in the xy-plane with radius 3, centered at the origin (0, 0, 0).
The Vector Field F and its Curl
The vector field F is given by F = -x i - y j, where i and j are the standard unit vectors in the x and y directions, respectively. The curl of F, denoted as curl F, is a vector field that measures the rotation or circulation of F.
The curl of F can be calculated using the formula: curl F = (∂Fz/∂y - ∂Fy/∂z) i + (∂Fx/∂z - ∂Fz/∂x) j + (∂Fy/∂x - ∂Fx/∂y) k, where i, j, and k are the standard unit vectors in the x, y, and z directions, respectively.
In this case, F = -x i - y j, so ∂Fz/∂y = 0, ∂Fy/∂z = 0, ∂Fx/∂z = 0, ∂Fz/∂x = 0, ∂Fy/∂x = -1, and ∂Fx/∂y = -1. Therefore, curl F = -1 k.
The Line Integral of F along the Closed Curve C
The line integral of a vector field F along a closed curve C can be calculated using the formula: ∮ F · dr, where dr is a differential vector along the curve C.
In this case, we need to calculate the line integral of F along the closed curve C that forms the boundary of the surface S. Since the surface S is a cone, the curve C is a circle in the xy-plane with radius 3, centered at the origin (0, 0, 0).
Since the vector field F = -x i - y j, we can write F · dr = (-x i - y j) · (dx i + dy j). Simplifying this expression, we get F · dr = -x dx - y dy.
To calculate the line integral, we need to parameterize the curve C. Let's use polar coordinates. Let r(t) = 3 cos(t) i + 3 sin(t) j, where t ranges from 0 to 2π.
Now, dr = (dx/dt) dt i + (dy/dt) dt j. Since x = 3 cos(t) and y = 3 sin(t), we have dx/dt = -3 sin(t) and dy/dt = 3 cos(t).
Substituting these values into F · dr = -x dx - y dy, we
Let S be the surface of cone z=√x^2 y^2 bounded by the planes z=0 and ...
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