Triple Integrals Suppose we have a function f ( x, y, z ) defined on - PDF document

Triple Integrals Suppose we have a function f ( x, y, z ) defined on a solid S ⊂ R 3 and we can estimate some quantity by � n i =1 f ( x ∗ i , y ∗ i , z ∗ i ) ∆ V i , where we divide the solid into small, compact pieces of volume ∆ V i and choose a point ( x ∗ i , y ∗ i , z ∗ i ) in each piece. We call those Riemann Sums. If they approach a limit, we call the limit ��� the triple integral of f over the solid and denote it by S f ( x, y, z ) dV . Applications of Triple Integrals ��� • S dV gives the volume of the solid S . • If δ ( x, y, z ) is the density of the solid at the point ( x, y, z ), then ��� M = S δ ( x, y, z ) dV gives the mass of the solid. ��� • M yz = S xδ ( x, y, z ) dV is the moment about the yz -plane. ��� • M xz = S yδ ( x, y, z ) dV is the moment about the xz -plane. ��� • M xy = S zδ ( x, y, z ) dV is the moment about the xy -plane. • If ( x, y, z ) is the center of mass of the solid, then x = M yz M , y = M xz M , z = M xy M . Evaluating a Triple Integral as an Iterated Integral Suppose a solid S ⊂ R 3 can be described as { ( x, y, z ) | α ( x, y ) ≤ z ≤ β ( x, y ), ( x, y ) ∈ D} , where D ⊂ R 2 is a plane region. �� β ( x,y ) � ��� �� We can then evaluate S f ( x, y, z ) dS = α ( x,y ) f ( x, y, z ) dz dA . D � β ( x,y ) �� We may wish to write this in the form α ( x,y ) dz f ( x, y, z ). D dA If D is a Type I region of the form { ( x, y ) | γ ( x ) ≤ y ≤ δ ( x ), a ≤ x ≤ b } , we may iterate the double integral to get � b � δ ( x ) � β ( x,y ) ��� S f ( x, y, z ) dV = α ( x,y ) dz f ( x, y, z ). a dx γ ( x ) dy We may think of the solid as { ( x, y, z ) | α ( x, y ) ≤ z ≤ β ( x, y ), γ ( x ) ≤ y ≤ δ ( x ), a ≤ x ≤ b } . Using Cylindrical Coordinates to Calculate Triple Integrals Consider a small solid obtained by starting at a point ( r, θ, z ) and letting each of the coordinates increase by ∆ r , ∆ θ and ∆ z . We get a solid which is almost a rectangular solid with vertices ( r, θ, z ), ( r + ∆ r, θ, z ), ( r, θ + ∆ θ, z ), ( r, θ, z + ∆ z ), ( r + ∆ r, θ + ∆ θ, z ), ( r + ∆ r, θ, z + ∆ z ), ( r, θ + ∆ θ, z + ∆ z ) and ( r + ∆ r, θ + ∆ θ, z + ∆ z ). The edge from ( r, θ, z ) to ( r + ∆ r, θ, z ) will have length ∆ r . 1

2 The edge from ( r, θ, z ) to ( r, θ + ∆ θ, z ) will have length r ∆ θ since it is parallel to the arc of a sector of a circle in the rθ − plane with radius r and angle ∆ θ . The edge from ( r, θ, z ) to ( r, θ, z + ∆ z ) will have length ∆ z . Thus the volume of the small solid will be ∆ V ≈ (∆ r )( r ∆ θ )(∆ z ) = r ∆ r ∆ θ ∆ z . Cylindrical Coordinates A sum � n i )∆ V i will thus ≈ � n i =1 f ( r ∗ i , θ ∗ i , z ∗ i =1 f ( r ∗ i , θ ∗ i , z ∗ i ) r ∗ i ∆ r i ∆ θ i ∆ z i and we may conclude ��� ��� S f ( r, θ, z ) dV = S f ( r, θ, z ) r dr dθ dz . Using Spherical Coordinates to Evaluate Triple Integrals Consider a small solid obtained by starting at a point ( ρ, θ, φ ) and letting each of the coordinates increase by ∆ ρ , ∆ θ and ∆ φ . We get a solid which is almost a rectangular solid with vertices ( ρ, θ, φ ), ( ρ + ∆ ρ, θ, φ ), ( ρ, θ + ∆ θ, φ ), ( ρ, θ, φ + ∆ φ ), ( ρ + ∆ ρ, θ + ∆ θ, φ ), ( ρ + ∆ ρ, θ, φ + ∆ φ ), ( ρ, θ + ∆ θ, φ + ∆ φ ) and ( ρ + ∆ ρ, θ + ∆ θ, φ + ∆ φ ). Spherical Coordinates The edge from ( ρ, θ, φ ) to ( ρ + ∆ ρ, θ, φ ) will have length ∆ ρ . The edge from ( ρ, θ, φ ) to ( ρ, θ +∆ θ, φ ) will have length r ∆ θ , where r is the corresponding cylindrical coordinate of the point. This was shown when looking at cylindrical coordinates. Since r = ρ sin φ , the edge will have length ρ sin φ ∆ θ . The edge from ( ρ, θ, φ ) to ( ρ, θ, φ +∆ φ ) will have length ρ ∆ φ , since it’s an arc subtended by an angle ∆ φ in a circle of radius ρ . Thus the volume of the small solid will be ∆ V ≈ (∆ ρ )( ρ sin φ ∆ θ )( ρ ∆ φ ) = ρ 2 sin φ ∆ ρ ∆ θ ∆ φ . A sum � n i =1 f ( ρ ∗ i , θ ∗ i , φ ∗ i )∆ V i will thus be i ) 2 sin( φ ∗ ≈ � n i =1 f ( ρ ∗ i , θ ∗ i , φ ∗ i )( ρ ∗ i )∆ ρ i ∆ θ i ∆ φ i and we may conclude S f ( ρ, θ, φ ) ρ 2 sin φ dρ dθ dφ . ��� ��� S f ( ρ, θ, φ ) dV = Change of Variable Using polar, cylindrical or spherical coordinates are special cases of a more general technique of a transformation, mapping or change of variables. It is a generalization of a change of variable for an ordinary integral. Consider a transformation T ( u, v ) = ( x, y ), where x = g ( u, v ), y = h ( u, v ) for some functions g : R 2 → R and h : R 2 → R , which associates

3 with every point ( u, v ) in some region S a corresponding point ( x, y ) in a region R . Assume the transformation is 1 − 1, onto and C 1 , meaning that g and h have continuous first-order partial derivatives. Suppose we take a small rectangle with vertices ( u 0 , v 0 ), ( u 0 + ∆ u, v 0 ), ( u 0 , v 0 + ∆ v ), ( u 0 + ∆ u, v 0 + ∆ v ). It will map into a region which is almost a parallelogram. Let T ( u 0 , v 0 ) = ( x 0 , y 0 ). T ( u 0 + ∆ u, v 0 ) = ( f ( u 0 + ∆ u, v 0 ) , g ( u 0 + ∆ u, v 0 )). Using differentials, f ( u 0 + ∆ u, v 0 ) ≈ x 0 + ∂x ∂u ∆ u . Similarly, g ( u 0 + ∆ u, v 0 ) ≈ y 0 + ∂x ∂u ∆ u . Effectively, one side of the parallelogram is the vector α = < ∂x ∂u ∆ u, ∂y ∂u ∆ u > = < ∂x ∂u, ∂y ∂u > ∆ u . Similarly, the other side of the parallelogram is effectively the vector β = < ∂x ∂v ∆ v, ∂y ∂v ∆ v > = < ∂x ∂v , ∂y ∂v > ∆ v . The area of the parallelogram is | α || β | sin θ , where θ is the angle be- tween the vectors. If we embed the two vectors in R 3 by adding a third component of 0 to each, we can calculate that with the cross product. � � i j k � � � � ∂x ∂y < ∂x ∂u, ∂y ∂u, 0 > ∆ u × < ∂x ∂v , ∂y � � 0 ∂v, 0 > ∆ v = ∆ u ∆ v = � � ∂u ∂u � � ∂x ∂y � � 0 � � ∂v ∂v � � � � ∂x ∂y � � � � ∂u ∂u ∆ u ∆ v k . � � ∂x ∂y � � � � ∂v ∂v � � � ∂ ( x, y ) � ∆ u ∆ v , where ∂ ( x, y ) � � � The length of the cross product is thus ∂ ( u, v ) = � � ∂ ( u, v ) � � � ∂x ∂y � � � � ∂u ∂u is called the Jacobian of the transformation. � � ∂x ∂y � � � � ∂v ∂v � � Thus, the area of the image in the xy − plane of the rectangle in the � � ∂ ( x, y ) � � uv − plane is approximately � ∆ u ∆ v . � � ∂ ( u, v ) �

4 If we had a Riemann sum � n i =1 f ( x i , y i ) ∆ A i over the region R , we � ∂ ( x, y ) � could approximate it by � n � � i =1 f ( g ( u i , v i ) , h ( u i , v i )) � ∆ u i ∆ v i . � � ∂ ( u, v ) � � � ∂ ( x, y ) � � �� �� We conclude R f ( x, y ) dA = S f ( g ( u, v ) , h ( u, v )) � du dv , � � ∂ ( u, v ) � � � ∂ ( x, y ) �� � � which can be written as S f ( x ( u, v ) , y ( u, v )) � du dv . � � ∂ ( u, v ) � Change of Variables in Higher Dimensions In higher dimensions, the visualization is trickier but the analogous results hold. For example, in R 3 , we would get ��� R f ( x, y ) dV = � � ∂ ( x, y, z ) � � ��� S f ( x ( u, v, w ) , y ( u, v, w ) , z ( u, v, w )) � du dv dw , � � ∂ ( u, v, w ) � ∂x ∂y ∂z � � � � � � ∂u ∂u ∂u � � where ∂ ( x, y, z ) ∂x ∂y ∂z � � ∂ ( u, v, w ) = . � � ∂v ∂v ∂v � � ∂x ∂y ∂z � � � � � ∂w ∂w ∂w �