Math 212-Lecture Integration in cylindrical and spherical coordinates

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1 Math 22-Lecture Integration in cylindrical and spherical coordinates Cylindrical he Jacobian is J = (x, y, z) (r, θ, z) = cos θ r sin θ sin θ r cos θ = r. Hence, d rdrdθdz. If we draw a picture, we can see directly that dv is really rdrdθdz. Spherical he Jacobian is (x, y, z) J = (ρ, φ, θ) = sin φ cos θ ρ cos φ cos θ ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cos φ ρ sin φ = ρ2 sin φ. Hence, d ρ 2 sin φdρdφdθ. If we draw a picture, we can see clearly that this is true. Example: Find the centroid of the first octant portion of the ball x 2 +y 2 +z 2 a 2 using both cylindrical coordinates and spherical coordinates, assuming the density is uniform. Solution. In Cylindrical way: the sphere is r 2 + z 2 = a 2. Hence, we can have r a, θ π/2, z a 2 r 2. Due to the symmetry, we must have x = ȳ = z. hen, z = zδd δdv dv zdv = a π/2 a 2 r 2 rdzdθdr a π/2 a 2 r 2 zrdzdθdr

2 In Spherical way: the sphere is ρ = a. Hence, ρ a, φ π/2, θ π/2. zd ρ 2 sin φdθdφdρ. ρ cos φ ρ 2 sin φdθdφdρ = Example: Write out the region bounded by z = x 2 + y 2 and z = y in cylindrical coordinates. Solution. We have done this example before. In cylindrical, they are z = r 2 and z = r sin θ. he intersection is r = sin θ. he projection onto xy plane is a circle. For θ, we set r =, and see, π are two adjacent zeros. Hence, θ π, r sin θ, r 2 z r sin θ. Example: Set up the integral for the area inside the two circles r = and r = 2 sin θ. Set up the integral for the volume of the solid bounded by r =, r = 2 sin θ, z = y and the xy plane. Solution. For the area, A = R da. In polar, da = rdrdθ. We see that we must divide the integral into three pieces. = 2 sin θ. We find θ = π/6 and θ = 5π/6. Hence, A = π/6 2 sin θ rdrdθ + 5π/6 π/6 he volume is R (z 2 z )da = R yda. Hence, π/6 2 sin θ r sin θrdrdθ+ 5π/6 π/6 π 2 sin θ rdrdθ + rdrdθ. 5π/6 ρ 3 sin φ cos φdθdφdρ. π 2 sin θ r sin θrdrdθ+ r sin θrdrdθ. 5π/6 Example: Set up the integral for the volume bounded by x 2 +y 2 +z 2 = 4 and x 2 + y 2 2x =. 2

3 Solution. We use cylindrical coordinates. r 2 + z 2 = 4 and r 2 cos θ =. he region is determined by r 2 cos θ =. We set r = and have cos θ =. Hence, π/2 θ π/2. hen, r 2 cos θ. For z, we find 4 r 2 z 4 r 2. Note d rdrdθdz. he volume is therefore: d π/2 2 cos θ 4 r 2 π/2 4 r 2 rdzdrdθ Example: Set up the integral for the mass of the region contained in the sphere x 2 + y 2 + (z a) 2 = a 2 but below z = r with unit density. If we draw the picture, we see that the spherical coordinates are the best. Solution. z = r is just φ = π/4. he sphere is ρ 2 2aρ cos φ = or ρ = 2a cos φ. Hence, we have π/4 φ π/2, ρ 2a cos φ, θ < 2π. hen, m = π/2 2a cos φ 2π π/4 ρ 2 sin φdθdρdφ. Example: Consider the ice-cream cone above φ = π/6 but below ρ = 2a cos φ. Suppose the density is δ =. Set up the integrals for the total mass and centroid. Solution. his problem is convenient in spherical coordinates. φ π/6, θ < 2π, ρ 2a cos φ. he volume element is d ρ 2 sin φdρdθdφ. he total mass is π/6 2π 2a cos φ m = δd ρ 2 sin φdρdθdφ. For the centroid, we use symmetry do conclude that x = ȳ =. hen, z = m because z = ρ cos φ. δzd m π/6 2π 2a cos φ ρ cos φρ 2 sin φdρdθdφ 3

4 4.8 Surface Area Previously, we see that a vector valued function with a single variable r(t) is a curve in space. Now, if the function is vector-valued but has two variables (parameters) r(u, v) = x(u, v), y(u, v), z(u, v), where r again is the position vector of the point, what will the object be? Fixing v = v, r(u, v ) is a curve. Now, for different v = v, it s another curve. he object is thus a family of curves, and they form a surface. Example: he graph z = f(x, y) is a surface in 3D space. Parametrize it. r = x, y, z. We choose x, y as the parameters. hen, r = x, y, f(x, y). Example: Let ρ, φ, θ be the spherical coordinates. he function ρ = h(φ, θ) gives a surface in the space. (Example is ρ = 2.) Parametrize this surface. Solution. r = x, y, z = h(φ, θ) sin φ cos θ, h(φ, θ) sin φ sin θ, h(φ, θ) cos φ. Example: Parametrize the rectangle x 2, y 3, z =. Solution. hence his is a special case of the first example, z = f(x, y) = and r(x, y) = x, y,, x 2, y 3 Given a parametric surface r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k = x(u, v), y(u, v), z(u, v), we call it smooth if r u = x u, y u, z u, r v = x v, y v, z v, are both nonzero and nonparallel. Consider the small area for the rectangle u v in u-v plane. It s a parallelogram on the surface under the mapping r(u, v). Draw a picture. One edge is a = r(u + u, v) r(u, v) r u u. he other edge is similarly b r v v. he area is therefore S a b r u r v u v. 4

5 N = r u r v is a normal vector of the surface. he total area is A = a(s) = ds = r u r v dudv. u,v u,v ds = r u r v dudv is the surface area element and ds = N N ds = Ndudv = r u r v dudv is the directed surface area element. Here, we see that r u r v plays the same role as the Jacobian in the change of variables for double integrals. It s the amplification factor between the areas. Example: If u, v are the Cartesian coordinates x, y, then r = x, y, f(x, y). It is the graph of z = f(x, y). r x r y = f x. f y,. his makes sense as it is just F where F = z f(x, y). he area is A = R + f 2 x + f 2 y dxdy. We compute the area of the ellipse cut from z = 2x + 2y + by x 2 + y 2 =. What if the surface is the one cut from x = 2y + 2z + from y + z =, y =, z =? Example: Find the area of the spiral ramp z = θ, r, θ π. Solution. We parametrize the surface r(r, θ) = r cos θ, r sin θ, θ. r r r θ = cos θ, sin θ, r sin θ, r cos θ, he magnitude of this is + r 2 he integral is π π/4 + r 2 dθdr = π sec 3 θdθ = π 2 ( 2 + ln( + 2)) Exercise. Compute the area of the portion of z 2 = 3(x 2 + y 2 ) below z = 3, and above xy plane. r = r cos θ, r sin θ, 3r. r 3, θ < 2π. he magnitude of r r r θ is 2r. hen, 3 2π 2rdθdr 5