Slide_13 - MA1104 Multivariable Calculus Lecture 13 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 13 Dr KU Cheng Yeaw Monday March 9, 2009 Recap Last lecture ... • evaluate double integrals over more general polar region • introduce triple integral over rectangular box, Type 1, Type 2 and Type 3 regions. Overview In this lecture, • we evaluate triple integral using Cylindrical Coordinates • we evaluate triple integral using Spherical Coordinates Triple Integrals in Cylindrical Coordinates We have seen that for a double integral over a region D , sometimes it is easier to compute the integral by changing our description of D to that using polar coordinate system in plane geometry. In three dimensions, there is a coordinate system, called cylindrical coordinates , that is similar to polar coordinates and gives convenient description of certain solid. We shall see how some triple integrals are much easier to compute using cylindrical coordinates. In the cylindrical coordinate system, a point P in three-dimensional (3-D) space is represented by the ordered triple ( r,θ,z ) , where: • r and θ are polar coordinates of the projection of P onto the xy plane. • z is the directed distance from the xy-plane to P . To convert from cylindrical coordinate to rectangular coordinates, we use the equations: x = r cos θ, y = r sin θ, z = z To convert from rectangular to cylindrical coordinates, we use the equations: r 2 = x 2 + y 2 , tan θ = y x , z = z Example 1. Describe the surface whose equation in cylindrical coordinates is z = r . Solution. Notice θ doesn’t appear in the equation. From the equation r 2 = x 2 + y 2 , we have z 2 = x 2 + y 2 which is the equation of a cone. Example 2. Describe the surface whose equation in cylindrical coordinates is r = c , where c is a constant. Solution. Using the equation x 2 + y 2 = r 2 , we have x 2 + y 2 = c 2 which is the equation of a cylinder. Suppose that E is a Type 1 region whose projection D on the xy-plane is conveniently described in polar coordinates. In particular, suppose that f is continuous and E = { ( x,y,z ) | ( x,y ) ∈ D,u 1 ( x,y ) = z = u 2 ( x,y ) } where D is given in polar coordinates by: D = { ( r,θ ) | α ≤ θ ≤ β,h 1 ( θ ) ≤ r ≤ h 2 ( θ ) } . From our previous study, we know that ZZZ E f ( x,y,z ) dV = ZZ D " Z u 2 ( x,y ) u 1 ( x,y ) f ( x,y,z ) dz # dA....
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Slide_13 - MA1104 Multivariable Calculus Lecture 13 Dr KU...

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