Math 235 - Review for Exam 2
1. Compute the second degree Taylor polynomial of f = e2x+3y about (0, 0).
Solution. A computation shows that
fx (0, 0) = 2, fy (0, 0) = 3, fxx (0, 0) = 4, fyy (0, 0) = 9, fxy (0, 0) = 6.
The second degree Taylor polynomial is then
f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1 2
2
fxx (0, 0)x + 2fxy (0, 0)xy + fyy (0, 0)y
2
= 1 + 2x + 3y + 2x + 6xy +
2
9 2
y .
2
2. Compute the second degree Taylor polynomials about (0, 0) of the following functions:
(a) f (x, y) = e2x−3y .
2
(b) f (x, y) = ex
−y 3
.
Solution: (a) There are two ways to do this one. One way is to use the definition of Taylor’s formula. A computation shows that
f (0, 0) = 1, fx (0, 0) = 2, fy (0, 0) = −3, fxx (0, 0) = 4, fxy (0, 0) = −6, fyy (0, 0) = 9.
The second degree Taylor polynomial is
P (x, y)
=
f (0, 0) + fx (0, 0)x + fy (0, 0)y +
=
1 + 2x − 3y + 2x − 6xy +
2
1
2
2
fxx (0, 0)x + fxy (0, 0)xy +
1
2
fyy (0, 0)y
2
9 2
y
2
Another way is to use the fact that eu ≈ 1 + u + u2 /2. Now let u = 2x − 3y to get
2x−3y
e
≈ 1 + (2x − 3y) +
1
2
(2x − 3y)
2
2
= 1 + x − 3y + 2x − 6xy +
9 2
y .
2
(b) Computing the Taylor polynomial using the formula will be a prohibitive computation. Use the trick in part (a). Let u = x2 − y 3 and
note that eu ≈ 1 + u + u2 /2 and so
1 2
x2 −y 3
2
3
3 2
e
≈ 1 + (x − y ) + (x − y ) .
2
Eliminate all the terms which are of degree 3 or higher (since we are looking for the second degree Taylor polynomial!) to get 1 + x2 as the second
degree Taylor polynomial.
3. Compute the second degree Taylor polynomial for f (x, y) = xsin(y) about (1, π2 ).
Solution: fx = sin(y) = 1, fy = xcos(y) = 0, fxx = 0, fyy = −xsin(y) = −1, fxy = cos(y) = 0
1 (y − π )2
So the Taylor polynomial is: 1 + 1(x − 1) − 2
2
4. For each of the functions below, find and identify (as max/max/saddle) all the critical points.
(a) f = 4xy − x4 − 2y 2
(b) f = x4 − y 4 .
Solution. (a) The critical points lie where the gradient is equal to zero. A computation shows that
3
∇f (x, y) = (4y − 4x , 4x − 4y) = (0, 0)
which generates the following system of equations:
3
4y − 4x
= 0, 4x − 4y = 0.
Using the second equation, we see that x = y. Plugging this into the first equation says x − x3 = 0 which occurs when x = 0, 1, −1. This the
critical points are (0, 0), (1, 1), (−1, −1). The Hessian turns out to be
−12x2
4
Hf (x, y) =
.
4
−4
Notice that
Hf (0, 0) =
0
4
1
4
−4
2
which has eigenvalues 2(−1 −
√
√
5) ≈ −6.47214 and 2(−1 + 5) ≈ 2.47214, yielding that (0, 0) is a saddle point. Notice that
Hf (1, 1) =
−12
4
4
−4
√
√
which has eigenvalues 4(−2 − 2) ≈ −13.6569 and 4(−2 + 2) ≈ −2.34315, yielding that (1, 1) is a maximum. Finally, notice that Hf (−1, −1) =
Hf (1, 1) and so (−1, −1) is also a maximum.
(b) A computation shows that the only critical point is (0, 0). Another computation shows that Hf (0, 0) is the zero matrix and so the second
derivative test fails. Notice that f (x, 0) = x4 which is always increases as you move away from zero in the x-direction while f (0, y) = −y 4 which
is always decreasing as you move away from zero in the y-direction. Thus (0, 0) is a saddle point.
5. Find the critical points of f (x, y) = x2 + 5xy − y 3 and classify them as local maximum, local minimum,
or saddle points.
Solution. The critical points (x, y) of f lie where ∇f (x, y) = (0, 0). Here
2
∇f (x, y) = (2x + 5y, 5x − 3y ) = (0, 0)
when (x, y) is a solution of the system
5x − 3y
2x + 5y = 0
2
= 0.
Solving this system (solving for x in the first equation and using that to solve for y in the second equation) gives the two critical points (0, 0) and
(125/12, −25/6). The Hessian of f is equal to
2
5
fxx
fxy
Hf (x, y) =
=
.
fyx
fyy
5
−6y
Examining the point (0, 0) we find that
2
5
Hf (0, 0) =
which has eigenvalues 1 −
(125/12, −25/6) we get
√
26 and 1 +
√
26.
5
0
These eigenvalues are of opposite signs and so (0, 0) is a saddle point.
For the critical point
2
5
Hf (125/12, −25/6) =
5
25
√
√
1 (23 + 829) ≈ 25.8962 and 1 (23 − 829) ≈ 2.89618. They are both positive sign and so the critical point (125/12, −25/6)
which has eigenvalues 2
2
is a minimum.
6. Find the critical points of f (x, y) = 8xy−0.25(x+y)4 and classify them as local maximum, local minimum,
or saddle points.
Solution. The critical points (x, y) of f lie where ∇f (x, y) = (0, 0). Here
3
3
∇f (x, y) = (8y − (x + y) , 8x − (x + y) ) = (0, 0)
when (x, y) is a solution of the system
8y − (x + y)
3
8x − (x + y)
=0
3
= 0.
3
Solving both of these equations for the quantity (x + y) , and setting them equal, we find that x = y. The first equation then becomes
8x = (x + x)3 = 8x3 , i.e., x = x3 which has solutions x = 0, 1, −1. Thus, the three critical points are (0, 0), (1, 1), (−1, −1). The Hessian of f is
equal to
fxx
fxy
−3(x + y)2
8 − 3(x + y)2
Hf (x, y) =
=
.
fyx
fyy
8 − 3(x + y)2
−3(x + y)2
For the critical point (0, 0),
Hf (0, 0) =
0
8
8
0
which has eigenvalues 8 and −8. Since they are of opposite sign, (0, 0) is a saddle point. For the critical point (1, 1),
Hf (1, 1) =
−12
−4
−4
−12
which has eigenvalues −16 and −8 which are both negative and so (1, 1) is a maximum. For the critical point (−1, −1),
Hf (−1, −1) =
−12
−4
−4
−12
which has eigenvalues −16 and −8, making the critical point (−1, −1) a maximum.
7. Find and classify the critical points of the function
f (x, y) =
x2
+ 3y 3 + 9y 2 − 3xy + 9y − 9x.
2
3
Solution. To find the critical points we set the gradient equal to zero and solve for (x, y). In this case,
fx = −9 + x − 3y = 0
fy = 9 − 3x + 18y + 9y
2
= 0.
Solving for x in the first equation and inserting it into the second yields the quadratic equation
2
9y + 9y − 18 = 0
which has the solutions y = −2 and y = 1. This gives us the two critical points
(3, −2)
and
(12, 1).
The Hessian (at a general point (x, y) is equal to
Hf =
fxx
fyx
fxy
fyy
=
−3
18y + 18
1
−3
.
1
−3
−3
−18
√
1 (−17 −
Which has eigenvalues λ = 2
397) ≈ −18.4624 and λ = 1
(−17 + 397) ≈ 1.46243. Thus (3, −2) is a saddle point.
2
1
−3
Hf (12, 1) =
−3
36
√
√
1 (37 −
which has eigenvalues λ = 2
1261) ≈ 0.744719 and λ = 1
(37 + 1261) ≈ 36.2553. Thus (12, 1) is a minimum.
2
Hf (3, −2) =
√
8. Find and classify the critical points of the function f (x, y) = x4 + y 4 .
Solution. Just as in the previous problem, the critical points are where the gradient is equal to zero. So
3
=0
3
=0
fx = 4x
fy = 4y
which holds when x = 0 and y = 0. Thus (0, 0) is the only critical point. The Hessian at a general point (x, y) is equal to
fxx
fxy
12x2
0
Hf =
=
fyx
fyy
0
12y 2
which, unfortunately, is the zero matrix at the critical point (0, 0) and so has both eigenvalues equal to zero. Thus the second derivative test is of
no use. However, notice that f (x, y) = x4 + y 4 is either equal to zero, at the critical point (0, 0), or it is positive. Hence (0, 0) is a local minimum.
9. Find and classify the critical points of the function f (x, y) = xy +
1
1
+ .
x y
Solution: The critical points are located at where ∇f = (0, 0). A computation shows that
2
2
∇f = (y − 1/x , x − 1/y ) = (0, 0)
2
1
when (x, y) = (1, 1). The Hessian at (1, 1) is equal to
which has eigenvalues λ = 1 and λ = 3, making the critical point (1, 1) a local
1
2
minimum.
10. For the function f (x, y) = x3 + x2 − y 3 + y 2 find all of the critical points and identify as local max., local
min., or saddle point.
Solution: fx = 3x2 + 2x = x(3x + 2) so fx = 0 implies x = 0 or x = − 2
.
3
.
fy = −3y 2 + 2y = y(−3y + 2) so fy = 0 implies y = 0 or y = 2
3
fxx = 6x + 2, fyy = −6y + 2, and fx y = 0, so D = (6x + 2)(−6y + 2)
D(0, 0) = 4 and fxx (0, 0) = 2. So there is a local min at (0, 0).
D(0, 2
) = −4. So there is a saddle at (0, 2
).
3
3
2 , 0) = −4. So there is a saddle at (− 2 , 0).
D(− 3
3
, 2 ) = 4, and fxx (− 2
, 2 ) = −2. So there is a local max at (− 2
, 2 ).
D(− 2
3 3
3 3
3 3
11. Compute the following integral:
Z
0
1
Z
1
√
y
p
2 + x3 dx dy
4
Solution. Reverse the order of integration
to get the above integral equal to
Z 1 Z x2 p
2 + x3 dy dx
0
0
which can be computed as
Z 1 Z x2 p
2 + x3 dy dx
0
=
Z 1
p
2
x
2 + x3 dx
0
0
=
≈
2
3/2
(3
−2
9
0.526161
3/2
12. Evaluate the following integral
Z
1
Z
1
√
0
p
x3 + 1 dx dy.
y
Solution. The region of integration is
Reversing the order of integration from dx dy to dy dx converts the above integral to
√
Z 1 Z x2 p
2
4 2
x3 + 1 dy dx = − +
.
9
9
0
0
13. Evaluate the integral
Z
e
Z
1/y
cos(x − log x)dx dy
1
Solution. Reverse the order of integration
1/e
)
5
to transform the integral to
Z e Z 1/y
Z 1 Z 1/x
cos(x − log x)dx dy
1
cos(x − log x)dy dx
=
1/e
1/e
1
Z 1
cos(x − log x)(
=
1/e
1
x
− 1)dx
x=1
=
− sin(x − log x)|x=1/e
=
− sin(1) + sin(1/e + 1)
14. Evaluate the integral
1
Z
e
Z
ey
0
x
dx dy.
log x
Solution. Reverse the order of integration
to transform the integral to
Z 1Z e
ey
0
Z e Z log x
x
log x
dx dy
=
1
Z e
=
x
dy dx
y=log x
log x
1
x
log x
0
y|y=0
dx
Z e
=
xdx
1
=
1
2
2
(e − 1)
15. Evaluate the following integral
Z
0
2
Z
1
2
ex dx dy.
y/2
Solution. This integral cannot be done as is since the x integral is impossible. So, reverse the order of integration
to transform the integral to
Z 1 Z 2x
x2
e dx dy
0
=
0
Z 1 Z 2x
x2
e dy dx
0
0
Z 1
=
0
=
x2 y=2x
|y=0 dx
ye
e−1
6
16. Evaluate the integral
8
Z
Z
2
x1/3
0
y4
1
dy dx.
+1
Solution. Reverse the order of integration
to transform the integral to
Z 2 Z y3
0
0
1
y4 + 1
dx dy
which can now be evaluated easily to log(17)/4.
Z
2
Z
4
17. Compute
0
y2
sin x
√ dx dy
x
Solution: Reverse the order of integration to get
Z 2Z 4
sin x
√ dx dy
x
0
y2
=
Z 4 Z √x
sin x
dy dx
√
x
0
0
Z 4
√
sin x y= x
y √
dx
x y=0
0
Z 4
sin xdx
=
1 − cos 4
=
=
0
18. Evaluate the following integral
√
Z
2
0
Z √4−y2
2
ex
+y 2
dx dy.
y
Solution. For this integral we draw the region of integration
and switch to polar coordinates to transform the integral to
Z π/4 Z 2
re
0
0
r2
dr dθ
7
which can easily be evaluated to π(e4 − 1)/8.
19. Evaluate
√
1/ 2
Z
Z √1−y2
0
sin(x2 + y 2 ) dx dy
y
Solution. Draw the region of integration
and switch to polar coordinates to get the above integral is equal to
Z π/4 Z 1
2
sin(r )r dr dθ
=
0
0
=
=
≈
20. Evaluate
√
Z
27/2
√
Z
9−x2
√
0
x/ 3
π
Z 1
4
π
0
4
π
r sin r
−
1
2
2
dr
2 r=1 cos(r ) r=0
1
1
−
cos(1)
4
2
2
0.180523
1
p
dy dx.
2
x + y2
Solution. Draw the region of integration
and switch to polar coordinates (note that
and that
p
√
√
y = x/ 3 = 9 − x2 ⇒ x = 27/3
√
1
1
y = x/ 3 ⇒ r sin θ = r cos θ √ ⇒ tan θ = √ ⇒ θ = π/6)
3
3
to transform the integral to
Z π/2 Z 3
dr dθ = π.
π/6
0
21. Find the area between the two curves x2 + y 2 = y and x2 + y 2 = x.
Solution. By completing the square, the above two curves are
2
2
= y ⇒ x + (y − 1/2)
2
2
= x ⇒ (x − 1/2) + y
x +y
x +y
Draw the region of integration
2
2
2
= 1/4 ⇒ circle: center (0, 1/2), radius 1/2
2
= 1/4 ⇒ circle: center (1/2, 0), radius 1/2
8
Now integrate in polar coordinates. First convert the two circles to polar coordinates
2
2
= x ⇒ r = cos θ
2
2
= y ⇒ r = sin θ.
x +y
x +y
Now use the fact that the region is symmetric about the line y = x so find the total area by taking twice the area of the ”bottom” part of the
region. This area is
Z π/4 Z sin θ
2
rdr dθ
0
0
which evaluates to π/8 − 1/4.
22. Set up
Z
f dA
R
as an integral, over the following region, in polar coordinates
Soln: This circle is x2 + (y − 2)2 = 4 which, making the substitutions x = r cos θ, y = r sin θ, becomes the circle r = 4 sin θ, 0 ≤ θ ≤ π. Thus
Z
Z π Z 4 sin θ
f dA =
f r dr dθ.
R
0
0
23. Set up
Z
f dA
R
as an integral, over the following region, in polar coordinates
Soln: This circle is (x + 2)2 + y 2 = 4 which, making the substitutions x = r cos θ, y = r sin θ, becomes the circle r = −4 cos θ, π/2 ≤ θ ≤ 3π/2.
Thus
Z
Z 3π/2 Z −4 cos θ
f dA =
f r dr dθ.
R
π/2
0
9
24. Set up
Z
f dA
R
as an integral, over the following region, in polar coordinates
Soln: The inner circle is r = 2 cos θ and the outer circle is r = 4 cos θ, −π/2 ≤ θ ≤ π/2. Thus
Z π/2 Z 4 cos θ
Z
f dA =
f r dr dθ.
−π/2
R
2 cos θ
25. Set
p up but do not evaluate a triple integral which represents the volume of the solid which lies: under
z = x2 + y 2 , above the xy plane, inside x2 + y 2 = 2y.
Solution. The region of integration is
and so the volume (in cylindrical coordinates) is
Z π Z 2 sin θ Z r
r dz dr dθ.
0
0
26. ForZ each
integral draw the region of integration:
1 Z z Z y+z
(a)
f (x, y, z) dx dy dz
Z0 π4 Z0 π2 0Z 2
(b)
ρ2 sin φ dρ dφ dθ
0
0
1
0
10
Solution:
27. Set up each of the following integrals in any coordinate system you choose. Do not compute the integrals.
(a) The integral of f (x, y, z) = x2 over the region bounded by the plane 2x+4y +z = 6 and the coordinate
planes.
(b) The integral f (x, y, z) = x2 + y 2 + z 2 over the region bounded by the cylinder x2 + z 2 = 2 , the plane
y = x and the plane y = 5.
2
2
2
2
(c) The integral of f (x, y, z) = 1 over the
p region bounded by the cylinders x + y = 1 and x + y = 2
2
2
2
2
and the surfaces z = x + y − 3 and z = 3 − x − y .
Solution: (a)
Z 3 Z 3−2x Z 6−2x−4y
2
2
x dz dy dx
0
0
0
q
Z √2 Z 5 Z 2−x2
2
2
2
q
(b)
x + y + z dz dy dx
√
−
2
x
−
2−x2
q
Z 2π Z √2 Z 3−r2
(c)
0
1
r 2 −3
r dz dr dθ
28. Let ρ(x, y, z) = x2 + y 2 + z 2 be a density function for the solid bounded by the surfaces z =
and z = 2. Find the mass of the solid.
p
x2 + y 2
Solution. The solid is drawn as follows:
The mass is the integral over the solid of the density function. Writing the integral in cylindrical coordinates gives us the mass is equal to
Z 2π Z 2 Z 2
2
2
(r + z )rdzdrdθ =
0
0
r
48π
5
11
p
2
2
29. Set
p up the volume integral of the function f (x, y) = x + y over the solid bounded by the surfaces
z = x2 + y 2 and z = 1 in three ways: rectangular coordinates, cylindrical coordinates, and spherical
coordinates. Then choose one of these three methods and compute the integral.
Solution. The region looks like
a. Rectangular:
−1
q
1−x2
q
− 1−x2
Z 1 Z
Z 1
q
x2 +y 2
q
x2 + y 2 dz dy dz
b. Cylindrical:
Z 2π Z 1 Z 1
2
r dz dr dθ = π/6
0
0
r
c. Spherical:
Z 2π Z π/4 Z sec φ
3
2
ρ sin φ dρ dφ dθ
0
0
0
30. Compute the volume enclosed by the surfaces
p
z = x2 + y 2 and x2 + y 2 = 2x.
Solution. The first surface is a cone and the second surface (after completing the square is (x − 1)2 + y 2 = 1 which is a cylinder centered at
(1, 0, 0) and parallel to the z-axis.
The shadow region in the xy plane is the circle x2 + y 2 = 2x which in polar coordinates is r = 2 cos θ. The volume is (in cylindrical coordinates)
Z π/2 Z 2 cos θ Z r
Z Z Z
dV
=
r dz dr dθ
−π/2
=
=
8
3
32
9
0
Z π/2
0
3
cos θ dθ
−π/2
12
31. Write down the integral which represents the volume of the region bounded by the surfaces x − y + z =
1, 12 x − 12 y + z = 1, y = x − 1, y = 0, x = 0.
Solution. The region of integration is
The shadow region is the region in the xy plane bounded by the x and y axis and the line y = x − 1. The volume is then
Z 1− 1 x+ 1 y
2
2
Z 1Z 0
0
x−1
dz dy dx.
1−x+y
32. Evaluate the following integral
Z Z Z
1
p
x2
+ y2
dz dy dx
over the region bounded by the surfaces x2 + y 2 = y, z = 0, z = y.
Solution. The region looks like
To write this integral in cylindrical coordinates, we first write all the bounding surfaces in cylindrical coordinates
2
x +y
2
= y ⇒ r = sin θ
z=0⇒z=0
z = y ⇒ z = r sin θ.
We now write the integral as
Z π Z sin θ Z r sin θ
1
0
0
0
r
r dz dr dθ
which, after a computation, can be evaluated to 2/3.
33. Evaluate the following integral
Z Z Z
(x2 + y 2 + z 2 ) dx dy dz
over the region bounded by the surfaces x2 + y 2 + z 2 = 2z and x2 + y 2 + z 2 = z.
Solution. The region of integration is the region contained between two spheres
13
Note that some completing the square is needed here to determine the center and radius of each sphere. Since we plan to integrate in spherical
coordinates, we write each sphere in spherical coordinates
2
2
x +y +z
2
2
2
x +y +z
= 2z ⇒ ρ = 2 cos φ
2
= z ⇒ ρ = cos φ
The integral (in spherical coordinates becomes)
Z 2π Z π/2 Z 2 cos φ
31π
2 2
ρ ρ sin φ dρ dφ dθ =
.
15
0
0
cos φ
34. Sketch the region of integration for the following triple integral
Z 6 Z 3− 12 x Z 6−x−2y
f (x, y, z) dz dy dx
0
0
0
Solution.
35. Sketch the region of integration for the following triple integral
Z 1 Z √1−x2 Z √2−x2 −y2
f (x, y, z) dz dy dx
−1
0
0
Solution.
36. Evaluate the integral
Z Z Z
(x2 + y 2 ) dz dy dx,
W
where W is the solid bounded by z = 0, z = 4, y = x, y = −x, y ≥ 0, x2 + y 2 = 2.
14
Solution. The region looks like
In cylindrical coordinates, the integral becomes
Z 3π/4 Z √2 Z 4
2
r r dz dr dθ = 2π
π/4
0
0
37. Draw, in detail and identify all surfaces involved, the solid whose volume is represented by the following
integrals:
Z π/6 Z 2π Z sec φ
(a)
ρ2 sin φ dρ dθ dφ
0
Z
0
2π
π/2
Z
0
Z
4 cos φ
ρ2 sin φ dρ dφ dθ
(b)
0
0
2 cos φ
Solution. (a)
(b)
38. Convert each integral below into an integral in rectangular coordinates. Do not actually compute the
integrals.
Z π/2 Z π/2 Z 5
(a)
ρ3 cos φ sin φ dρ dφ dθ
0
0
0
15
π
Z
√
Z
3
Z
3−r 2
r2 dz dr dθ
(b)
0
0
0
Solution: (a) The region of integration is the portion of the sphere x2 + y 2 + z 2 = 25 with x, y, z ≥ 0. Also note that
3
2
ρ cos φ sin φdρdφdθ = ρ cos φ(ρ sin φ)dρdφdθ = zdxdydz.
Finally,
Z π/2 Z π/2 Z 5
Z 5Z
3
ρ cos φ sin φ dρ dφ dθ =
0
0
0
q
25−x2
25−x2 −y 2
zdzdydx.
0
0
q
Z
0
(b) The region of integration is the solid bounded by the paraboloid z = 3 − x2 − y 2 and the plane z = 0 and with y ≥ 0. Also notice that
q
2
r dzdrdθ = r · rdzdrdθ = x2 + y 2 dzdxdy.
Finally,
q
Z π Z √3 Z 3−r2
Z √3 Z 3−x2 Z 3−x2 −y2 q
2
r dz dr dθ =
x2 + y 2 dzdydx.
√
0
0
−
0
3
0
0
2
2
39.
Z πThe volume enclosed by z = 1 − x − y and z = 1 − y can be written as an integral of the form
c
sind θ dθ. Find the constants c and d.
0
Solution: The region of integration is bounded below by the plane z = 1 − y and above by the paraboloid z = 1 − x2 − y 2 . The shadow region is
formed when 1 − x2 − y 2 = 1 − y, equivalently when x2 + y 2 = y which is the circle with radius 1/2 and center (0, 1/2). In polar coordinates this
circle is r = sin θ (use x2 + y 2 = y with x = r cos θ and y = r sin θ). The volume is then
Z π Z sin θ Z 1−r2
V
=
r dzdrdθ
0
0
1−r sin θ
Z π Z sin θ
=
0
z=1−r2
rz z=1−r sin θ
0
Z π Z sin θ
=
2
(−r + r sin θ) dr dθ
0
=
3
dr dθ
0
Z π
−r 4
0
Z π
=
0
4
1
12
+
r3
3
r=sin θ
sin θ dθ
r=0
4
sin θ dθ
So c = 1/12 and d = 4.
p
40. Give three integrals for the volume inside the cone x2 + y 2 ≤ z ≤ 1, one in rectangular coordinates,
one in cylindrical coordinates, and the third in spherical coordinates. Do not evaluate any of the integrals.
Solution: q
Z 1 Z 1−x2 Z 1
−1
−
q
1−x2
q
x2 +y 2
1 dz dy dx
Z 2π Z 1 Z 1
r dz dr dθ
0
0
r
Z 2π Z π Z sec(φ)
4
2
ρ sin(φ) dρ dφ dθ
0
0
0
p
41. Compute the value of the integral of 1/ x2 + y 2 + z 2 over the region between the spheres of radius 1
and 2 centered at the origin.
Solution:
Z 2 Z π Z 2π
1
1
0
0
ρ
2
ρ sin(φ) dρ dφ dθ = 6π