Vextendsingle vextendsingle vextendsingle

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Unformatted text preview: vextendsingle vextendsingle vextendsingle vextendsingle dt . 2.2 Line integral of a vector field In the physical world, we often use the integral integraldisplay C vector F · d vector r . We can write d vector r as ˆ T ds where ˆ T is the unit (normalized) tangential vector along the curve: ˆ T = d vector r dt vextendsingle vextendsingle vextendsingle vextendsingle d vector r dt vextendsingle vextendsingle vextendsingle vextendsingle , because ˆ T ds = d vector r dt vextendsingle vextendsingle vextendsingle vextendsingle d vector r dt vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle d vector r dt vextendsingle vextendsingle vextendsingle vextendsingle dt = d vector r . Therefore, integraldisplay C vector F · d vector r = integraldisplay C vector F · ˆ T ds . This shows us that the interpretation of this integral is that we are integrating the com- ponent of the vector field that is tangential to the curve, i.e. the component of the field that lies along the curve. Also, this is really an integration of a scalar quantity (or scalar field, vector F · ˆ T ) along the curve. 2.3 Surface integral of a scalar field A parametric surface defined by a pointing vector vector r ( u , v ) = ( r 1 ( u , v ) , r 2 ( u , v ) , r 3 ( u , v ) ) defines the coordinates of the points at which the integration occurs. For a scalar field f ( x , y , z ) , the scalar field values at these points are f ( r 1 ( u , v ) , r 2 ( u , v ) , r 3 ( u , v )) because r 1 ( u , v ) , r 2 ( u , v ) , and r 3 ( u , v ) are respectively the x , y , and z components of the pointing vector that points to the surface. The infinitesimal area on the surface is dS = vextendsingle vextendsingle vextendsingle vextendsingle ∂ vector r ∂ u × ∂ vector r ∂ v vextendsingle vextendsingle vextendsingle vextendsingle du dv . Using the expressions above, the integral of a scalar field on the parametric surface is integraldisplayintegraldisplay S f ( x , y , z ) dS = integraldisplayintegraldisplay f ( r 1 ( u , v ) , r 2 ( u , v ) , r 3 ( u , v )) vextendsingle vextendsingle vextendsingle vextendsingle ∂ vector r ∂ u × ∂ vector r ∂ v vextendsingle vextendsingle vextendsingle vextendsingle du dv . EE 505 C, Autumn, 2010 Vector Calculus 11 2.4 Surface integral of a vector field In the physical world, we often use the integral integraldisplayintegraldisplay S vector F · d vector S , where d vector S = parenleftbigg ∂ vector r ∂ u × ∂ vector r ∂ v parenrightbigg du dv . d vector S has magnitude that is an infinitesimal area, and direction that is normal to the surface. We can write d vector S as ˆ n dS where ˆ n is the unit (normalized) tangential vector along the curve: ˆ n = ∂ vector r ∂ u × ∂ vector r ∂ v vextendsingle vextendsingle vextendsingle vextendsingle ∂ vector r ∂ u × ∂ vector r ∂ v vextendsingle vextendsingle vextendsingle vextendsingle , because ˆ n dS = ∂ vector r ∂ u × ∂ vector...
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