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TRIPLE INTEGRALS IN SPHERICAL COORDINATES

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1 TRIPLE INTEGRALS IN SPHERICAL COORDINATES
SECTION 12.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES

2 SPHERICAL COORDINATE SYSTEM
Another useful coordinate system in three dimensions is the spherical coordinate system. It simplifies the evaluation of triple integrals over regions bounded by spheres or cones. 12.7

3 SPHERICAL COORDINATES
The spherical coordinates (r, , f) of a point P in space are shown in Figure 1. r = |OP| is the distance from the origin to P.  is the same angle as in cylindrical coordinates. f is the angle between the positive z-axis and the line segment OP. 12.7

4 SPHERICAL COORDINATES
Note that: r ≥ 0 0 ≤  ≤ π 12.7

5 SPHERICAL COORDINATE SYSTEM
The spherical coordinate system is especially useful in problems where there is symmetry about a point, and the origin is placed at this point. 12.7

6 SPHERE For example, the sphere with center the origin and radius c has the simple equation r = c. This is the reason for the name “spherical” coordinates. See Figure 2. 12.7

7 The graph of the equation  = c is a vertical half-plane.
See Figure 3. 12.7

8 The equation f = c represents a half-cone with the z-axis as its axis.
See Figure 4. 12.7

9 SPHERICAL & RECTANGULAR COORDINATES
The relationship between rectangular and spherical coordinates can be seen from Figure 5. 12.7

10 SPHERICAL & RECTANGULAR COORDINATES
From triangles OPQ and OPP’, we have: z = r cos f r = r sin f However, x = r cos  y = r sin  12.7

11 SPH. & RECT. COORDINATES So, to convert from spherical to rectangular coordinates, we use the equations x = r sin f cos  y = r sin f sin  z = r cos f 12.7

12 Also, the distance formula shows that: r 2 = x2 + y2 + z2
SPH. & RECT. COORDINATES Also, the distance formula shows that: r 2 = x2 + y2 + z2 We use this equation in converting from rectangular to spherical coordinates. 12.7

13 The point (2, p/4, p /3) is given in spherical coordinates.
Example 1 The point (2, p/4, p /3) is given in spherical coordinates. Plot the point and find its rectangular coordinates. 12.7

14 We plot the point in Figure 6.
Example 1 SOLUTION We plot the point in Figure 6. 12.7

15 From Equations 1, we have:
Example 1 SOLUTION From Equations 1, we have: 12.7

16 Thus, the point (2, p/4, p /3) is in rectangular coordinates.
Example 1 SOLUTION Thus, the point (2, p/4, p /3) is in rectangular coordinates. 12.7

17 The point is given in rectangular coordinates.
Example 2 The point is given in rectangular coordinates. Find spherical coordinates for the point. 12.7

18 Example 2 SOLUTION From Equation 2, we have: 12.7

19 Therefore, spherical coordinates of the given point are:
Example 2 SOLUTION So, Equations 1 give: Note that  ≠ 3p/2 because Therefore, spherical coordinates of the given point are: (4, p/2 , 2p/3) 12.7

20 EVALUATING TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
In the spherical coordinate system, the counterpart of a rectangular box is a spherical wedge where a ≥ 0, b – a ≤ 2p, d – c ≤ p 12.7

21 EVALUATING TRIPLE INTEGRALS
Although we defined triple integrals by dividing solids into small boxes, it can be shown that dividing a solid into small spherical wedges always gives the same result. 12.7

22 EVALUATING TRIPLE INTEGRALS
Thus, we divide E into smaller spherical wedges Eijk by means of equally spaced spheres r = ri, half-planes  = j, and half-cones f = fk. 12.7

23 EVALUATING TRIPLE INTEGRALS
Figure 7 shows that Eijk is approximately a rectangular box with dimensions Δr : ri Δfk (arc of a circle with radius ri, angle Δfk) ri sinfkΔj (arc of a circle with radius ri sin fk, angle Δj) 12.7

24 EVALUATING TRIPLE INTEGRALS
So, an approximation to the volume of Eijk is given by: 12.7

25 EVALUATING TRIPLE INTEGRALS
In fact, it can be shown, with the aid of the Mean Value Theorem, that the volume of Eijk is given exactly by: where is some point in Eijk. 12.7

26 EVALUATING TRIPLE INTEGRALS
Let be the rectangular coordinates of that point. 12.7

27 EVALUATING TRIPLE INTEGRALS
Then, 12.7

28 EVALUATING TRIPLE INTEGRALS
However, that sum is a Riemann sum for the function So, we have arrived at the following formula for triple integration in spherical coordinates. 12.7

29 where E is a spherical wedge given by:
Formula 3 where E is a spherical wedge given by: 12.7

30 TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
Formula 3 says that we convert a triple integral from rectangular coordinates to spherical coordinates by writing: x = r sin f cos  y = r sin f sin  z = r cos f 12.7

31 TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
That is done by: Using the appropriate limits of integration. Replacing dV by r2 sinf dr d df. This is illustrated in Figure 8. 12.7

32 TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
The formula can be extended to include more general spherical regions such as: The formula is the same as in Formula 3 except that the limits of integration for r are g1(, f) and g2(, f). 12.7

33 TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
Usually, spherical coordinates are used in triple integrals when surfaces such as cones and spheres form the boundary of the region of integration. 12.7

34 Evaluate where B is the unit ball
Example 3 Evaluate where B is the unit ball 12.7

35 As the boundary of B is a sphere, we use spherical coordinates:
Example 3 SOLUTION As the boundary of B is a sphere, we use spherical coordinates: In addition, spherical coordinates are appropriate because: x2 + y2 + z2 = r2 12.7

36 Example 3 SOLUTION So, Formula 3 gives: 12.7

37 Note It would have been extremely awkward to evaluate the integral in Example 3 without spherical coordinates. In rectangular coordinates, the iterated integral would have been 12.7

38 Use spherical coordinates to find the volume of the solid that lies:
Example 4 Use spherical coordinates to find the volume of the solid that lies: Above the cone Below the sphere x2 + y2 + z2 = z See Figure 9. 12.7

39 Example 4 SOLUTION Notice that the sphere passes through the origin and has center (0, 0, ½). We write its equation in spherical coordinates as: r2 = r cos f or r = cos f 12.7

40 The equation of the cone can be written as:
Example 4 SOLUTION The equation of the cone can be written as: This gives: sin f = cos f or f = p/4 12.7

41 Thus, the description of the solid E in spherical coordinates is:
Example 4 SOLUTION Thus, the description of the solid E in spherical coordinates is: 12.7

42 Example 4 SOLUTION Figure 11 shows how E is swept out if we integrate first with respect to r, then f, and then . 12.7

43 Example 4 SOLUTION The volume of E is: 12.7

44 TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
Figure 10 gives another look (this time drawn by Maple) at the solid of Example 4. 12.7


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