Begin2.DVI
Generalization of Lagrange Multipliers In general, to find an extremal value associated with a n-dimensional function given by f ...
critical points. Whenever the determinant of the Hessian matrix is zero at a critical point, then the critical point ( ̄x 0 , ̄λ ...
are defined and illustrated. Throughout the following discussion all surfaces are considered to be oriented (two-sided) surfaces ...
If the surface is expressed in an implicit form F(x, y, z ) = 0 , then a unit normal to the surface can be obtained from the rel ...
Example 7-25. The parametric equations x=acos usin v, y =asin usin v, z =acos v with 0 ≤u≤ 2 π and 0 ≤v≤π, represent the surface ...
It can be verified that a unit normal to this surface is ˆen= N |N| = 1 ar. That is, the unit outer normal to a point P on th ...
Example 7-27. Consider a surface described by the implicit form F(x, y, z) = 0. Recall that equations of this form define zas a ...
The equation of the line through the point (x 0 , y 0 , z 0 )which is perpendicular to the tangent plane is given by r =r 0 +λ ...
Figure 7-19. Element of surface area and element parallelogram. In the limit as ∆xand ∆ytend toward zero, ∆Rapproaches ∆Sand one ...
Given a surface in the explicit form z=z(x, y ),define the outward normal to the surface φ(x, y, z ) = z−z(x, y ) = 0 by eˆn= gr ...
Example 7-28. Find the surface area of that part of the plane φ(x, y, z ) = 2x+ 2y+z−12 = 0 which lies in the first octant. Figu ...
and the limits of summation are determined as dx and dz range over the region x > 0 , z > 0 ,and 2 x+z≤ 12 .This produces ...
can be thought of as defining anelementofvolume dV in the shape of a parallelepiped with vector sides A =hudu ˆe 1 ,B =hvdv ˆe ...
represents a scalar which is the sum of the projections of F onto the normals to the surface elements. If the surface is divide ...
Example 7-29. Evaluate the surface integral ∫∫ R F·dS, where Sis the surface of the cube bounded by the planes x= 0, x = 1, y ...
(iii) On face BFGC the unit normal to the surface is the vector n =ˆe 2 and yhas the value 1 everywhere so that ∫∫ R F·dS= ∫ ...
The element of surface area dS is projected upon the xy plane giving dS = dx dy |ˆe 3 ·eˆn| = 3 dx dy Figure 7-22. Plane 2 x+ 2y ...
Example 7-31. Evaluate the surface integral ∫∫ R F×dS, where Sis the plane 2 x+ 2 y+z−1 = 0 in the first octant and F =xˆe 1 ...
Summary When a surface is represented in parametric form, the position vector of a point on the surface can be represented as r ...
One finds ∣∣ ∣∂r∂θ ×∂r∂φ ∣∣ ∣=r^2 sin θso that a unit vector to the surface of the sphere is ˆen= sin θcos φeˆ 1 + sin θsin φˆ ...
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