#$&* course Mth 277 4/19 3 uestion: `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): I did not reduce but understand what you did and why. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Convert the equation 3x^2 + 3y^2 + 3z^2 = 1 to spherical coordinates. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have the equation of a sphere were x^2+y^2+z^2 = (1/3) so using our formual Because `rho = `sqrt(x^2+y^2+z^2), then for our equation 3(x^2+y^2+z^2) =3* `rho^2, so we can write 3*`rho^2 = a^2, where `rho = (x^2+y^2+z^2) and a =1, so 3*`rho^2 = 1^2 = 1 `rho^2 = 1/3 Where the coordinates are from rectangular to spherical (x,y,z) = (`sqrt(x^2+y^2+z^2), (y/x), cos^-1(z/(`sqrt(x^2+y^2+z^2))) 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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Convert the equation rho = sin(theta)*cos(phi) to rectangular coordinates. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This equation seems to be in spherical coordinates, with `rho = sin(`theta)*cos(`phi), knowing that x = (`rho)*sin(`theta)*cos(`phi) we find x by x = (`rho)*(sin(`theta)*cos(`phi)) = (sin(`theta)*cos(`phi))*(sin(`theta)*cos(`phi)) =( sin(`theta)*cos(`phi))^2 = sin(`theta)^2*cos(`phi)^2 In the same manner we find y by y = (`rho)*(sin(`theta)*sin(`phi)) = (sin(`theta)*cos(`phi))*( sin(`theta)*sin(`phi)) = sin(`theta)^2*sin(`phi)*cos(`phi) Finally z is found by z = (`rho)*(cos(`theta)) = (sin(`theta)*cos(`phi))*(cos(`theta) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I understand the solution provided, my only question is I guess is that I get confused with what the question is looking for. For example when the question asks for something to be converted to “”whatever”” coordinates then wouldn’t our final solution be in the form (x,y,z) or (`rho, `theta,`phi) or if we were to answer the question by providing an equation, would our solution look like the one provided to this question? When I worked the problems in the book it stated “Rewrite the given equations in “whatever” form” and the solutions looked similar to the one provided in this question. This is a very minor misunderstanding and nothing that would keep me from being able to understand or complete this work. Just want to make sure I understand what you want to see me work out. I do understand the solutions provided and can take that info and find the and see what our solution would be in coordinate form. ------------------------------------------------ Self-critique rating:
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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): ------------------------------------------------ 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: It’s not stated but I believe that I am to answer this in spherical coordinates because we have a region which is a sphere(and that’s what the section is on). If so we have the equation in rectanglular form with r = `sqrt(5) We know that V=Int(Int(Int( dV(over S))) = Int(Int(Int(`rho^2*sin(`theta)*`drho,0,r) `dtheta,0,`pi)`dphi,0,2`pi) Finding our values of the spherical coordinates `rho = `sqrt(5), note this is the same value as the radius we found in rec. equation ???From the info given we can’t find anything else right, except `rho. Which is all we need to know to solve this triple integral???? No we have Int(`rho^2*sin(`theta)*`drho,0,r) = sin(`theta)*(`rho^3/3), evaluated sin(`theta)*(`sqrt(5)^3/3) = sin(`theta)(5/3)*`sqrt(5) Int(sin(`theta)(5/3)*`sqrt(5) `dtheta,0,`pi) = (5/3)*`sqrt(5)(-cos(`theta)), evaluated (5/3)*`sqrt(5)[ -cos(`pi) - (-cos(0))] = (5/3)*`sqrt(5)[1 + 1] = (5/3)*`sqrt(5)*2 = (10/3)`sqrt(5) Int((10/3)*`sqrt(5)`dphi,0,2`pi)= (10/3)`sqrt(5)Int( `dphi,0,2`pi)= (10/3)`sqrt(5)*`phi, evaluated (10/3)`sqrt(5)[2`pi - 0] = (10/3)`sqrt(5)*2`pi = (20`pi*`sqrt(5))/3), which is equal to = approx. 46.832 When we use the formual for Volume of a Sphere we get (4/3)*`pi*r^3 = (4/3)*`pi*(`sqrt(5))^3 =approx. 46.832 Where both solutions to Integral and formula are rounded , they are still exactly the same solutions. Which it is not hard to see why when logic of the triple integral is followed and understanding what goes on during the volume change for sphere. 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): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!