<p> 1</p><p>Section 9.7/12.7: Triple Integrals in Cylindrical and Spherical Coordinates</p><p>Practice HW from Larson Textbook (not to hand in) Section 9.7: p. 606 # 1-35 odd Section 12.7: p. 780 # 1-4 , Supplemental Exercises at end of notes # 1-8</p><p>Cylindrical Coordinates</p><p>Cylindrical coordinates extend polar coordinates to 3D space. In the cylindrical coordinate system, a point P in 3D space is represented by the ordered triple (r,, z) . Here, r represents the distance from the origin to the projection of the point P onto the x- y plane,  is the angle in radians from the x axis to the projection of the point on the x-y plane, and z is the distance from the x-y plane to the point P. z</p><p>P(r,, z)</p><p> y r </p><p> x</p><p>As a review, the next page gives a review of the sine, cosine, and tangent functions at basic angle values and the sign of each in their respective quadrants. 2</p><p>Sine and Cosine of Basic Angle Values</p><p>  cos sin sin Degrees Radians tan  cos 0 0 cos0  1 sin 0  0 0 30  3 1 3 6 2 2 3 45  2 2 1 4 2 2 60  1 3 3 3 2 2 90  0 1 undefined 2 180  -1 0 0 270 3 0 -1 undefined 2 360 2 1 0 0</p><p>Signs of Basic Trig Functions in Respective Quadrants</p><p> cos sin sin Quadrant tan  cos I + + + II - + - III - - + IV + - -</p><p>The following represent the conversion equations from cylindrical to rectangular coordinates and vice versa.</p><p>Conversion Formulas</p><p>To convert from cylindrical coordinates (r,, z) to rectangular form (x, y, z) and vise versa, we use the following conversion equations.</p><p>From cylindrical to rectangular form: x  r cos , y  r sin , z = z.</p><p> y From rectangular to cylindrical form: r 2  x2  y 2 , tan  , and z = z x 3</p><p>Example 1: Convert the points ( 2, 2,3) and (3, 3,1) from rectangular to cylindrical coordinates.</p><p>Solution: 4</p><p>█ 5</p><p> Example 2: Convert the point (3, ,1) from cylindrical to rectangular coordinates. 4</p><p>Solution:</p><p>█</p><p>Graphing in Cylindrical Coordinates</p><p>Cylindrical coordinates are good for graphing surfaces of revolution where the z axis is the axis of symmetry. One method for graphing a cylindrical equation is to convert the equation and graph the resulting 3D surface. 6</p><p>Example 3: Identify and make a rough sketch of the equation z  r 2 .</p><p>Solution:</p><p> z</p><p> y</p><p> x █</p><p> Example 4: Identify and make a rough sketch of the equation   . 4</p><p>Solution:</p><p> z</p><p> y</p><p>█ x 7</p><p>Spherical Coordinates</p><p>Spherical coordinates represents points from a spherical “global” perspective. They are good for graphing surfaces in space that have a point or center of symmetry.</p><p>Points in spherical coordinates are represented by the ordered triple</p><p>(,,) where  is the distance from the point to the origin O,  , where is the angle in radians from the x axis to the projection of the point on the x-y plane (same as cylindrical  coordinates), and  is the angle between the positive z axis and the line segment OP joining the origin and the point P (,,) . Note 0     .</p><p> z</p><p> P(,,)</p><p></p><p> y</p><p></p><p> x 8</p><p>Conversion Formulas</p><p>To convert from spherical coordinates (,,) to rectangular form (x, y, z) and vise versa, we use the following conversion equations.</p><p>From spherical to rectangular form: x   sin cos , y   sin sin , z   cos y From rectangular to spherical form:  2  x2  y 2  z 2 , tan  , and x z z   arccos( )  arccos( ) x2  y 2  z 2 </p><p>Example 5: Convert the points (1, 1, 1) and (3, 3,2 2) from rectangular to spherical coordinates.</p><p>Solution: 9</p><p>█ 10</p><p> Example 6: Convert the point (9, , ) from rectangular to spherical coordinates. 4</p><p>Solution:</p><p>█</p><p>Example 7: Convert the equation   2sec to rectangular coordinates.</p><p>Solution:</p><p>█ 11</p><p> Example 8: Convert the equation   to rectangular coordinates. 3</p><p> z Solution: For this problem, we use the equation   arccos( ) . If we take x2  y 2  z 2 the cosine of both sides of the this equation, this is equivalent to the equation</p><p> z cos  x2  y 2  z 2  Setting   gives 3  z cos  . 3 x2  y 2  z 2</p><p> 1 Since cos  , this gives 3 2</p><p>1 z  2 x 2  y 2  z 2 or</p><p> x2  y 2  z 2  2z</p><p>Hence, x2  y 2  z 2  2z is the equation in rectangular coordinates. Doing some algebra will help us see what type of graph this gives.</p><p>Squaring both sides gives</p><p> x2  y 2  z 2  (2z)2 2 2 2 2 x  y  z  4z The graph of x2  y 2  3z 2  0 x2  y 2  3z 2  0 is a cone shape half whose two parts be found by graphing the two equations  x2  y 2  z 2  2z . The graph of the top part, x2  y 2  z 2  2z , is displayed as follows on the next page.</p><p>(continued on next page) 12</p><p>█</p><p>Example 9: Convert the equation x2  y 2  z to cylindrical coordinates and spherical coordinates.</p><p>Solution: For cylindrical coordinates, we know that r 2  x2  y 2 . Hence, we have r 2  z or r   z</p><p>For spherical coordinates, we let x   sin cos , y   sin sin , and z   cos to obtain</p><p>( sin cos )2  ( sin sin )2   cos</p><p>We solve for  using the following steps:</p><p> 2 sin 2  cos2    2 sin 2  sin 2    cos (Square terms)  2 sin 2 (cos2   sin 2  )   cos (Factor  2 sin 2 )  2 sin 2  (1)   cos  0 (Use identity cos2   sin 2   1) ( sin 2   cos)  0 (Factor )   0,  sin 2   cos  0 (Set each factor equal to zero and solve) cos   0,   sin 2  █ Triple Integrals in Cylindrical Coordinates 13</p><p>Suppose we are given a continuous function of three variables f (r,, z) expressed over a solid region E in 3D where we use the cylindrical coordinate system.</p><p> z</p><p> z  h2 (r, )</p><p>E y r  g2 ( ) z  h (r, ) 1 1 2</p><p> r  g1( ) x</p><p>Then</p><p> 2 rg2 ( ) zh2 (r, )  f (r,, z) dV     f (r,, z) r dz dr d E  1 rg1( ) zh1(r, )</p><p> 2 rg2 ( ) zh2 (r, ) Volume of E   dV     r dz dr d E  1 rg1( ) zh1(r, ) 14</p><p>3 2 Example 10: Use cylindrical coordinates to evaluate  (x  xy ) dV , where E is the E solid in the first octant that lies beneath the paraboloid z  1 x2  y 2 .</p><p>Solution: 15</p><p>█ 16</p><p>Example 11: Use cylindrical coordinates to find the volume of the solid that lies both within the cylinder x2  y 2  4 and the sphere x2  y 2  z 2  9 .</p><p>Solution: Using Maple, we can produce the following graph that represents this solid:</p><p>In this graph, the shaft of the solid is represented by the cylinder equation x2  y 2  4 . It is capped on the top and bottom by the sphere x2  y 2  z 2  9 . Solving for z, the upper and bottom portions of the sphere can be represented by the equations z   9  x2  y 2 . Thus, z ranges from z   9  x2  y 2 to z  9  x2  y 2 . Since x2  y 2  r 2 in cylindrical coordinates, these limits become z   9  r 2 to z  9  r 2 .When this surface is projected onto the x-y plane, it is represented by the circle x2  y 2  4 . The graph is</p><p>(Continued on next page) 17</p><p>This is a circle of radius 2. Thus, in cylindrical coordinates, this circle can be represented from r = 0 to r = 2 and from   0 to   2 . Thus, the volume can be represented by the following integral: 2  2 rg2 ( ) zh2 (r, )  2 r2 z 9r Volume   dV     r dz dr d     r dz dr d E  1 rg1( ) zh1(r, )  0 r0 z 9r2</p><p>We evaluate this integral as follows: 2  2 r2 z 9r2  2 r2 z 9r    r dz dr d    rz dr d 2  0 r0 z 9r2  0 r0 z 9r  2 r2    r( 9  r 2 )  r( 9  r 2 ) dr d  0 r0  2 r2    2r 9  r 2 dr d  0 r0 r2  2 3 2 2   (9  r 2 ) d (Use u - du sub let u  9 - r 2 )  3  0 r0  2 3 3 2 2 2 2  [ (9  22 )   (9  02 ) ] d  3 3  0  2 3 3 2 2 2 2  [ (5)  (9) ] d  3 3  0  2 3 3 10 2 2  [18  5] d (Note (9)  27 and (5)  5 5)  3  0 10  2  [18  5] 3  0 10  (18  5)2  0 3 20  36  5 3</p><p>20 Thus, the volume is 36  5 . 3</p><p>█ 18</p><p>Triple Integrals in Spherical Coordinates</p><p>Suppose we have a continuous function f (,, ) defined on a bounded solid region E.</p><p> z</p><p> P(,,)</p><p></p><p> y</p><p></p><p> x</p><p>Then</p><p> 2 2 h2 (, )  f (,, ) dV     f (,, )  2sin d d d E  1 1 h1(, )</p><p> 2 2 h2 (, ) Volume of E   dV      2sin d d d E  1 1 h1(, ) 19</p><p> x 2  y 2  z 2 Example 12: Use spherical coordinates to evaluate  e dV , where Q is Q enclosed by the sphere x2  y 2  z 2  9 in the first octant.</p><p>Solution: 20</p><p>█ 21</p><p>2 2 2 16x2 16x  y Example 13: Convert    x 2  y 2 dz dy dx from rectangular to 2 0 0 spherical coordinates and evaluate.</p><p>Solution: Using the identities x   sin cos and y   sin sin , the integrand becomes </p><p> x2  y 2   2 sin 2  cos2    2 sin 2  sin 2 </p><p>  2 sin 2 (cos2   sin 2  )</p><p>  2 sin 2 (1)   sin</p><p>The limits with respect to z range from z = 0 to z  16  x2  y 2 . Note that z  16  x2  y 2 is a hemisphere and is the upper half of the sphere x2  y 2  z 2  16 . The limits with respect to y range from y = 0 to y  16  x2 , which is the semicircle located on the positive part of the y axis on the x-y plane of the circle x2  y 2  16 as x ranges from x  4 to x  4 . Hence, the region described by these limits is given by the following graph</p><p> Thus, we can see that  ranges from   0 to   4 ,  ranges from   0 to   and 2  ranges from   0 to    . Using these results, the integral can be evaluated in polar coordinates as follows:</p><p>(continued on next page) 22</p><p>2 2 4 16 x 2 16 x  y    x2  y2 dz dy dx 4 0 0      2  4      sin ( 2 sin) d d d  0  0  0      2  4      3 sin2  d d d  0  0  0       4 2  4  sin2  d d (Integrate with respect to  )   4  0  0  0         2 44   2  [ sin2   0]d d  64 sin2  d d (Sub in limits and simplify)   4    0  0  0  0      2 1  cos 2  1  cos 2u  64 d d (Use trig identity sin2 u  )    2  2  0  0           2   2    32(1  cos 2) d d    ( 32  32 cos 2 )d d (Simplify and dist 32)  0  0  0  0      1 2  ( 32  32( )sin 2 ) (Integrate with respect to , use u - du sub for cos2)  2  0  0        2     ( 32 16sin 2 ) d  [32   16sin 2( ) ]  (32(0) 16sin 0) d    2  2  0  0  0       (16 16sin  0)d   (16 16(0) )d  0  0    16 d  16   (Integrate with respect to  )   0  0  16 ( )  0  16 2 █ 23</p><p>Supplemental Exercises </p><p>2 2 1. Use cylindrical coordinates to evaluate  x  y dV , where Q is the region that Q lies inside the cylinder x2  y2  16 and between the planes z  5 and z  4 . (Answer 384 )</p><p> ez dV 2. Use cylindrical coordinates to evaluate  , where Q is the region enclosed by Q the paraboloid z  1  x2  y2 , cylinder x2  y2  5, and the x-y plane. (Answer  (e6  e  5) )</p><p>3. Find the volume of the region Q bounded by the paraboloids z  x2  y2 and z  36  3x2  3y2 . (Answer 162 )</p><p>(x2  y2  z2 ) dV 4. Use spherical coordinates to evaluate  , where Q is the unit ball Q 4 x2  y2  z2  1 . (Answer ) 5</p><p> z dV 5. Use spherical coordinates to evaluate  , where Q lies between the spheres Q 15 x2  y2  z2  1 and x2  y2  z2  4 in the first octant. (Answer ) 16</p><p> x2 dV 6. Use spherical coordinates to evaluate  , where Q is bounded by the xy-plane Q 1562 and the hemispheres z  9  x2  y2 and z  16  x2  y2 . (Answer ) 15</p><p>2 2 4 y 2 7. Evaluate the integral    xz dz dx dy by changing to cylindrical 2  4 y 2 x 2  y 2 coordinates. (Answer 0) 2 2 1 1 x 2 2 x  y 8. Evaluate the integral    xy dz dy dx by changing to cylindrical 0 0 x 2  y 2 coordinates. (Answer (4 2  5) /15)</p>