Math 21a Homework 24 Solutions Spring, 2014

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1 Math a Homework olutions pring, Due Friday, April th (MWF) or Tuesday, April 5th (TTh) This assignment is officially on urface Area (ection.6) and calar urface Integrals (ection.6), but it s most useful for practicing your parametrizations! You are encouraged to think about and draw the relevant surfaces and the grid lines of the parametrizations as practice.. Find the area of the given surface: (a) (tewart.6 #8 ) The helicoid (or spiral ramp) with vector equation r(u, v) = u cos v i + u sin v j + v k, and u and v π. (Note: the integral is doable, but very annoying. One strategy is: make an x = tan t substitution, and then integrate the result by parts with u = sec t. You must at least set up the integral correctly, but if you don t want to integrate it by hand, it s fine to use Mathematica s Integrate command (it works the same way as NIntegrate from HW ) to find an exact answer.) Recall that the surface area of this parmeterized surface is Area = π r u r v du dv. Here the partial derivatives are r u = cos v, sin v, and r v = u sin v, u cos v,, so their cross product is r u r v = sin v, cos v, u. This has length r u r v = u +. Thus the surface area is Area = π u + du dv = π u + dv du = π u + du. This is the integral described as very annoying in the problem statement. Mathematica gives us the answer + arcsinh() Area = π, and with a little investigation we can determine that arcsinh() = ln( + ). But how did we get there? Here we follow the hint in the problem statement and let u = tan(t), so du = sec (t) dt. The limits change from u = to to t = arctan() = to arctan() = π. We get Area = π π/ π/ tan t + sec t dt = π sec t dt (where we ve used the equality tan t + = sec t). Now this integral is a classic pain and we won t actually solve it except in that footnote. uffice it to say the answer is Area = π = π π/ sec t dt [ = π sec t tan t + [ ( ) = π + ( ) + ln( + ). This is the same as the answer from Mathematica. ] π/ ln sec t + tan t ( ln( + ) ln( + ) It has it s own Wikipedia page: Basically the idea is to integrate by parts with u = sec t and dv = sec t dt, so v = tan t. Then you get, after using the tan t + = sec t equality again, a formula like sec t dt = sec t tan t sec t dt + sec t dt. Then you use this the trick of adding sec t dt to both sides. This leaves us with the integral of sec t, which has it s own Wikipedia page: The general approach there is to know the answer already. This sounds like a joke, but really what you do is re-write sec t in a way that you only know to do because you ve seen the trick before: Combining these gives sec t dt = sec t + sec t tan t sec t + tan t dt = ln sec t + tan t + C. sec t dt = sec t tan t + ln sec t + tan t + C. ) ]

2 (b) (tewart.6 # ) The part of the paraboloid x = y + z that lies inside the cylinder y + z = 9. (This integral is quite doable by hand, and worth the practice.) In the non-standard cylindrical coordinates (x, r, θ) with x = x, y = r cos θ, z = r sin θ, the paraboloid can be parametrized by s(r, θ) = r, r cos θ, r sin θ θ < π, r. The partial derivatives are s r = r, cos θ, sin θ and s θ =, r sin θ, r cos θ, so their cross product is s r s θ = r, r cos θ, r sin θ and this has length s r s θ = 5 r. We are interested in the part of the paraboloid with r. The surface area of this part is s r s θ dr dθ = 5r dr dθ = π 5 r dr = 8π 5.. In this problem, we ll find the surface area of the part of the cylinder x +y = 9 that lies between the planes x+y+z = 6 and x + y + z = 5. (a) Use Mathematica to plot the cylinder and the planes: Plane = ContourPlotD[x+y+z == 5,{x,-,},{y,-,},{z,-,}] Plane = ContourPlotD[x+y+z == -6,{x,-,},{y,-,},{z,-,}] MyCylinder = ContourPlotD[x^+y^ == 9,{x,-,},{y,-,},{z,-,}] how[mycylinder, Plane,Plane] (b) Find a parametrization r (θ) of the curve C of intersection of the plane x+y +z = 5 with the cylinder x +y = 9. Also find a parametrization r (θ) of the curve C of intersection of the plane x + y + z = 6 with the cylinder. In both cases we parameterize x and y via x = cos θ, y = sin θ, then solve for z in terms of x and y, and thus in terms of θ. We get that C and C are parameterized by In both cases θ < π. r (θ) = cos θ, sin θ, 5 cos θ sin θ, r (θ) = cos θ, sin θ, 6 cos θ sin θ. (c) Write down a parametrization s(θ, u) of the part of the cylinder that lies between the two planes. The curves C and C should be two grid curves of the parametrization, and the bounds on the parameters should be of the form a u b and c θ d for constants a, b, c, d. One can take s(θ, u) = cos θ, sin θ, u cos θ sin θ with θ π and 6 u 5. (This is simply intersecting with the planes x + y + z = u from u = 6 to u = 5.)

3 (d) Use mathematica to plot your parametrization s(θ, u). (It s probably easiest to call the first parameter t instead of θ. You ll need to replace x(t,u), y(t,u) and z(t,u) by the components you found for s(t, u) in part (c), and the a,b,c,d by your bounds from part (c)): ParamCylinder = ParametricPlotD[{x(t,u),y(t,u),z(t,u)},{u,a,b},{t,c,d}] how[paramcylinder, Plane,Plane] (e) Use the parametrization you found to calculate the surface area of the part of the cylinder that lies between the two planes. The partial derivatives are s θ = sin θ, cos θ, sin θ cos θ and s u =,,. Hence s θ s u = cos θ, sin θ, =. The surface area is then equal to 5 s θ s u du dv = du dθ = π = 66π. (f) Does your answer from (e) make sense? Give an intuitive geometric reason that the surface area you found is the same as the surface area of a cylinder of radius and height. Every vertical segment that makes up the skewed cylinder has height. By sliding the segments so they all begin at a single height, we get a standard cylinder of height. This standard cylinder can be unwrapped to a cylinder of height and width 6π (the circumference of a circle of radius ), and thus this standard cylinder has surface area 66π.

4 . (Optional: 5 extra credit points) Evaluate the surface integral. (a) (tewart.6 #8 ) (x +y where is the surface with vector equation ) d r(u, v) = uv, u v, u + v, u + v (This integral is doable if you are careful about simplifying the integrand (and choosing a good coordinate system).) First let s simplify the integrand, as per the suggestion in the problem: x + y = (uv) + (u v ) = u v + u u v + v = u + u v + v = (u + v ). Then we compute r u r v in the usual way: r u = v, u, u, r v = u, v, v and so and Thus our surface integral is r u r u = 8uv, u v, u v r u r u = (8uv) + (u v ) + ( u v ) = 6u v + (6u u v + 6v ) + (6u + u v + 6v ) = u + 6u v + v = ( u + v ). (x + y ) d = u +v Changing to polar coordinates (u = r cos θ and v = r sin θ), we get (x + y ) d = (r ) (r ) r dr dθ = (u + v ) (u + v ) du dv. r 7 dr dθ = 8 π = π. (b) (tewart.6 # ) x +y +z d (Each of the integrals should be doable.) where is the part of the cylinder x + y = 9 between the planes z = and z =, together with the top and bottom disks. Note that has three parts to it: the sides of the cylinder which we ll call sides, the top top and the bottom bottom. We compute the integral separately for each piece. For sides, we use the cylindrical parametrization r(θ, z) = cos θ, sin θ, z with θ < π and z. ince the partial derivatives are r θ = sin θ, cos θ, and r z =,,, we see that the cross product is r θ r z = cos θ, sin θ, =. Therefore x + y + z d = sides (9 + z ) dz dθ = π [7z + z ] = π (5 + 8) = π. For bottom, we use the polar parameterization r(r, θ) = r cos(θ), r sin(θ), with r and θ < π. Then r r = cos(θ), sin(θ), and r θ = r sin(θ), r cos(θ),, so r r r θ =,, r and r r r θ = r. (This is just saying that in part of the xy-plane, the area element in polar coordinates is simply r dr dθ.) Thus bottom x + y + z d = r r dr dθ = r dr dθ = π = 8π.

5 Finally, for top, we use the polar parameterization r(r, θ) = r cos(θ), r sin(θ), with r and θ < π. Then r r and r θ are the same as with bottom, so we again get r r r θ = r. Hence top x + y + z d = (r + ) r dr dθ = ( ) (r + r) dr dθ = + π = 5π. Putting this all together, we get x + y + z d = x + y + z d + x + y + z d + x + y + z d sides bottom top = π + 8π + 5π = π.. (tewart.6 #8 ) Find the mass of a thin funnel in the shape of a cone z = x + y, z, if its density function is ρ(x, y, z) = z. We work in cylindrical coordinates, so our funnel is the graph of the function z(r, θ) = r. Thus we can parameterize this funnel as r(r, θ) = r cos(θ), r sin(θ), r ( θ < π and r ), so r r r θ = cos(θ), sin(θ), r sin(θ), r cos(θ), = r cos(θ), r sin(θ), r = r. Thus the funnel has mass equal to ( r) r dr dθ = π ] [5r r = π [( 8 6 ) ( 5 )] = 8π. 5. (Optional: extra credit points) (tewart.6 #6 ) The two cylinders y + z = and x + z = bound a solid region in R. The region is drawn on page 87. Find the surface area of the region. The region in question is made out of four identical pieces. One of them (the one that intersects the positive x-axis) can be parameterized by r(y, z) = z, y, z for y + z. Thus r y r z = z,,,, = z,, Therefore the surface area of this one piece is y +z dy dz = z z z = z z z dy dz. z. This is an improper integral (the denominator z = when z = ±), but we ll plow ahead and see what happens. The inner integral is straightforward, and we end up with something simple: dy dz = y y= z z z z dz = z dz = dz =. z y +z y= (Technically we should be taking limits as z approaches and to avoid the impropriety, but in the end we ll get for this integral.) ince this was the surface area of one of four identical pieces of the surface, the full surface has area () = 6. Reading: Please read ection. on vector fields, which we ll be discussing next class. If you are in the TTh section, please also skim through ection..