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a2 x2 dx
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As an aid in setting up this integral, note that z dy dx corresponds to the volume of a column such as shown shaded in the gure. Keeping x constant and integrating with respect to y from y 0 to p darkly y a2 x2 corresponds to adding the volumes of all such columns in a slab parallel to the yz plane, thus giving the volume of this slab. Finally, integrating with respect to x from x 0 to x a corresponds to adding the volumes of all such slabs in the region, thus giving the required volume.
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9.5. Find the volume of the region bounded by z x y; z 6; x 0; y 0; z 0
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MULTIPLE INTEGRALS
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Fig. 9-10
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Required volume volume of region shown in Fig. 9-10 6 6 x f6 x y g dy dx 6 6
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x 0 y 0
 1 6 x 6 x y y2  dx 2 y 0 x 0 1 6 x 2 dx 36 x 0 2
In this case the volume of a typical column (shown darkly shaded) corresponds to f6 x y g dy dx. The limits of integration are then obtained by integrating over the region r of the gure. Keeping x constant and integrating with respect to y from y 0 to y 6 x (obtained from z 6 and z x y corresponds to summing all columns in a slab parallel to the yz plane. Finally, integrating with respect to x from x 0 to x 6 corresponds to adding the volumes of all such slabs and gives the required volume.
TRANSFORMATION OF DOUBLE INTEGRALS 9.6. Justify equation (9), Page 211, for changing variables in a double integral.
In rectangular coordinates, the double integral of F x; y over the region r (shaded in Fig. 9-11) is F x; y dx dy. We can also evaluate this double integral by considering a grid formed by a family of u and v curvilinear coordinate curves constructed on the region r as shown in the gure.
Fig. 9-11
CHAP. 9]
MULTIPLE INTEGRALS
Let P be any point with coordinates x; y or u; v , where x f u; v and y g u; v . Then the vector r from O to P is given by r xi yj f u; v i g u; v j. The tangent vectors to the coordinate curves u c1 and v c2 , where c1 and c2 are constants, are @r=@v and @r=@u, respectively. Then the area of region r of    @r @r  Fig. 9-11 is given approximately by   u v. @u @v But     i j k    @x @y       @x @y    @r @r  @ x; y  @u @u 0   @u @u k k   @x @y   @u @v  @ u; v        @x @y @v @v 0  @v @v so that      @r @r      u v @ x; y  u v @u @v  @ u; v  X taken over the entire region r.   @ x; y   Ff f u; v ; g u; v g  @ u; v  u v
The double integral is the limit of the sum
An investigation reveals that this limit is   @ x; y   Ff f u; v ; g u; v g  @ u; v  du dv
where r 0 is the region in the uv plane into which the region r is mapped under the transformation x f u; v ; y g u; v . Another method of justifying the above method of change of variables makes use of line integrals and Green s theorem in the plane (see 10, Problem 10.32).
9.7. If u x2 y2 and v 2xy, nd @ x; y =@ u; v in terms of u and v.
 @ u; v  ux  @ x; y  vx
   uy   2x 2y     4 x2 y2 vy   2y 2x  p u2 v2
From the identify x2 y2 2 x2 y2 2 2xy 2 we have x2 y2 2 u2 v2 Then by Problem 6.43, 6, @ x; y 1 1 1 p @ u; v @ u; v =@ x; y 4 x2 y2 4 u2 v2 Another method: Solve the given equations for x and y in terms of u and v and nd the Jacobian directly. and x2 y2
9.8. Find the polar moment of inertia of the region in the xy plane bounded by x2 y2 1, x2 y2 9, xy 2; xy 4 assuming unit density.
Under the transformation x2 y2 u, 2xy v the required region r in the xy plane [shaded in Fig. 9-12(a)] is mapped into region r 0 of the uv plane [shaded in Fig. 9-12(b)]. Then:   @ x; y   x2 y2 dx dy x2 y2  Required polar moment of inertia  @ u; v  du dv p 8 du dv 1 9 u2 v2 p du dv 8 4 u2 v2 4 u 1 v 4 0
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