#$&* course Mth 277 December 3 around 1:40pm. Question: `q001. Convert the point (-3, 2pi/3, 3) from cylindrical coordinates to both rectangular and spherical coordinates. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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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): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q002. Convert the equation 3x^2 + 3y^2 + 3z^2 = 1 to spherical coordinates. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** I’m not sure if this will work but I’ll see: ro= sqrt (x^2+y^2+z^2) ro^2=( x^2+y^2+z^2) 3*ro^2=1 ro^2=1/3 sqrt (1/3)=0.577 #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): YES! So much easier!
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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): OK, so you take your “ro” and set that equal to your cos(phi)? Why do you set it equal to cos(phi) instead of sin(theta)?
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q005. Evaluate the triple integral of sqrt(x^2 + y^2 + z^2) with respect to V over R where R is the region defined by x^2 + y^2 + z^2 <= 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** I did it using SPHERICAL coordinates: Regions: ro<=sqrt (5) 0<=theta<=2pi 0<=phi<=pi/2 dV=ro^2*sin(phi) dro, dtheta, dphi Int (Int (Int ro*ro^2*sin(phi) dro), dtheta), dphi) with the above boundaries #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to give the integrand and should’ve simplified the integral to (ro^3*sin(phi)). I got everything else right though. ------------------------------------------------ self-critique rating #$&*: