If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
Question: `q001. Convert the point (-3, 2pi/3, 3) from cylindrical coordinates to both rectangular and spherical coordinates.
Your solution:
Confidence rating:
Given Solution:
The coordinates of this point are r = -3, theta = 2 pi / 3 and z = 3.
Quick solution:
x = r cos(theta) = 1.5
y = r sin(theta) = -3 sqrt(3) / 2
z is the same in both coordinate systems, equal to 3.
So the rectangular-coordinate representation of the point is (1.5, -3 sqrt(3) / 2, 3).
rho = sqrt(r^2 + z^2) = sqrt(3) * 2
phi = arcsin(r / rho) = arcsin(3 / (2 sqrt(3)) = pi/4 (alternatively phi = arcTan(z / r) = arcTan(3/3) = pi/4)
theta is unchanged
The spherical-coordinates representation is thus (3, 2 pi/3, pi/4).
Geometric solution with reasoning:
The point (r, theta) = (-3, 2 pi/3) lies in the x-y plane and its coordinates are
x = r cos(theta) = - 3 cos(2 pi/3) = -3 * -.5 = 1.5
and
y = r sin(theta) = -3 sin(2 pi / 3) = -3 * sqrt(3) / 2 = -3 sqrt(3) / 2, approximately -2.6.
The z coordinate is the given z = 3.
So in rectangular coordinates the point is (1.5, -3 sqrt(3) / 2, 3), approximately (1.5, -2.6, 3).
To convert to spherical coordinates (rho, theta, phi) we observe that a triangle with one leg running from the origin to the point (-3, 2 pi / 3) in the x-y plane, and another running from that point to the given point (-3, 2 pi/3, 3), is a right triangle whose hypotenuse runs along the radial line, between the origin and (-3, 2 pi/3, 3). Both legs have length 3 so the hypotenuse has length sqrt( 3^2 + 3^2 ) = 3 sqrt(2). This is the radial coordinate rho.
The angle phi between the radial line and the z axis is identical to the angle between the vertical leg of our triangle and the radial line (that vertical line and the z axis are parallel, and the radial line forms alternate interior angles between these parallel lines). Thus sin(phi) = 3 / (3 sqrt(2)) = sqrt(2) / 2, and phi = arcsin ( sqrt(2) / 2) = pi/4.
The spherical-coordinates representation is thus (3, 2 pi/3, pi/4).
Self-critique (if necessary):
Self-critique rating:
Question: `q002. Convert the equation 3x^2 + 3y^2 + 3z^2 = 1 to spherical coordinates.
Your solution:
Confidence rating:
Given Solution:
Easy way (geometric and algebraic insight):
By the Pythagorean Theorem rho^2 = (x^2 + y^2 + z^2) so
3 x^2 + 3 y^2 + 3 z^2 = 3 * rho^2, and our equation is
3 rho^2 = 1 or
rho^2 = 1/3.
Using the formulas:
x = rho sin(phi) cos(theta), y = rho sin(phi) sin(theta), z = rho cos(phi) so
3 x^2 + 3 y^2 + 3 z^2 = 3 rho^2 sin^2(phi) cos^2(theta) + 3 rho^2 sin^2(phi) sin^2(theta) + 3 rho^2 cos^2(phi)
= 3 rho^2 sin^2(phi) (cos^2(theta) + sin^2(theta) ) + 3 rho^2 cos^2(phi)
= 3 rho^2 sin^2(phi) * 1 + 3 rho^2 cos^2(phi)
= 3 rho^2 ( sin^2(phi) + cos^2(phi) )
= 3 rho^2 * 1
= 3 rho^2
so
3 rho^2 = 1 and
rho^2 = 1/3.
Self-critique (if necessary):
Self-critique rating:
Question: `q003. Convert the equation rho = sin(theta)*cos(phi) to rectangular coordinates.
Your solution:
Confidence rating:
Given Solution:
rho = sqrt(x^2 + y^2 + z^2)
sin(theta) = y / sqrt(x^2 + y^2)
cos(phi) = z / sqrt(x^2 + y^2 + z^2)
So our equation is
sqrt(x^2 + y^2 + z^2) = z / sqrt( x^2 + y^2 + z^2) so that
z = (x^2 + y^2 + z^2) and
x^2 + y^2 + z^2^ - z = 0.
(note that if we complete the square we get x^2 + y^2 + (z - 1/2)^2 = 1/4, the equation of a circle centered at (0, 0, 1/2) with radius 1/2)
Self-critique (if necessary):
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Question: `q004. Evaluate the iterated integral Int( Int( Int( r dz, 0, 4cos(theta)) dr, 0, sin(theta)) d(theta), 0, pi/2).
Your solution:
Confidence rating:
Given Solution:
The inner integral gives us antiderivative r z, which between z = 0 and z = 4 cos(theta) changes by 4 r cos(theta).
Integrating this next with respect to r we get antiderivative 2 r^2 cos(theta), which between theta = 0 and r = sin(theta) changes by 2 sin^2(theta) cos(theta).
Integrating this with respect to theta we let u = sin(theta) and get du = cos(theta) dTheta, which leads fairly directly to antiderivative 2 sin^3(theta) / 3. Between theta = 0 and theta = pi/2 this changes by 2/3.
Our result is thus 2/3.
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Question: `q005. Evaluate the triple integral of sqrt(x^2 + y^2 + z^2) with respect to V over R where R is the region defined by x^2 + y^2 + z^2 <= 5.
Your solution:
Confidence rating:
Given Solution:
The given region can be described in rectangular coordinates by
-sqrt(5) <= x <= sqrt(5), sqrt(5 - x^2) <= y <= sqrt( 5 - x^2), sqrt( 5 - x^2 - y^2) <= z <= sqrt(5 - x^2 - y^2),
in polar coordinates by
0 <= r <= sqrt(5), 0 <= theta <= 2 pi, 0 <= z <= sqrt( 5 - r^2)
or in spherical coordinates by
rho <= sqrt(5), 0 <= theta <= 2 pi, 0 <= phi <= pi/2.
The integrand is described in cylindrical coordinates by
sqrt(r^2 + z^2)
and in spherical coordinates by
sqrt(rho).
The volume increment for rectangular coordinates is dV = `dx * `dy * `dz.
The volume increment for cylindrical coordinates is dV = r `dr `dTheta.
The volume increment for spherical coordinates is dV = rho^2 sin(phi) `dRho `dTheta `dPhi
The rectangular-coordinate integral is
int(int(int(sqrt(x^2 + y^2 + z^2) dz, -sqrt(5 - x^2 - y^2), sqrt(5 - x^2 - y^2)) dy, -sqrt(5 - x^2), sqrt(5 - x^2)) dx, -sqrt(5), sqrt(5)).
This is a mess to evaluate (start with tan(u) = z / sqrt(x^2 + y^2) and go from there, if you wish).
In polar coordinates the integral is
int(int(int(sqrt(r^2 + z^2) dz, -sqrt(5 - r^2), sqrt(5 - r^2)) r dr, 0, sqrt(5)) dtheta, 0, 2 pi).
If you start with tan(u) = z / r, this isn't too bad, if you like trigonometric substitutions (which aren't bad and are frequently unavoidable).
In spherical coordinates the integral is
int(int(int(rho * rho^2 sin(phi) dRho, 0, sqrt(5)) dPhi, 0, pi) dTheta, 0, 2 pi),
which is a piece of cake (you get 25 pi).
Our three integrals, in reverse of the order presented above, are
However the three integrals do not appear to yield the same result. The spherical-coordinate integral and the cylindrical-coordinate integral do yield the correct result.
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