The following parametric equations describe a portion of a shperical surface:
x=4cosusinv
y=4sinusinv
z=4cosv
where 0<=u<=pi/4 and 0<=v<=pi/2
Find the surface area of this portion of the sphere.
To find the surface area of the portion of a spherical surface described by the given parametric equations, we can use the formula for the surface area of a surface given by parametric equations:
S = ∫∫ ∥r_u × r_v∥ dudv
where r(u, v) = <x(u, v), y(u, v), z(u, v)> represents the parametric equations of the surface, r_u represents the partial derivative of r with respect to u, r_v represents the partial derivative of r with respect to v, and ∥r_u × r_v∥ represents the magnitude of the cross product of r_u and r_v.
Let's start by finding the partial derivatives of r:
r_u = <x_u, y_u, z_u>
r_u = <(-4sinu*sinv), (4cosu*sinv), 0>
r_v = <x_v, y_v, z_v>
r_v = <(0), (0), (-4cosv)>
Next, let's calculate the cross product of r_u and r_v:
r_u × r_v = (det |i j k| ; |(-4sinu*sinv) (4cosu*sinv) 0| ; |0 0 -4cosv|)
= <(-16cosu*sin^2v), (-16sinu*sin^2v), (0)>
Now, find the magnitude of r_u × r_v:
∥r_u × r_v∥ = sqrt[(-16cosu*sin^2v)^2 + (-16sinu*sin^2v)^2 + (0)^2]
= sqrt[256cos^2u*sin^4v + 256sin^2u*sin^4v]
= sqrt[256sin^4v(cos^2u + sin^2u)]
= sqrt[256sin^4v]
Now, we can set up the double integral to calculate the surface area:
S = ∫∫ ∥r_u × r_v∥ dudv
= ∫[0, π/2]∫[0, π/4] sqrt[256sin^4v] dudv
To simplify the integral, we can use the fact that sin^2v = 1 - cos^2v:
S = ∫[0, π/2]∫[0, π/4] 16sin^2v du dv
= 16 ∫[0, π/2] sin^2v du dv
To evaluate this integral, we can use the trigonometric identity sin^2v = (1 - cos(2v))/2:
S = 16 ∫[0, π/2] [(1 - cos(2v))/2] du dv
= 8 ∫[0, π/2] (1 - cos(2v)) dv
Integrating each term separately:
8 ∫[0, π/2] (1 - cos(2v)) dv
= 8 [v - (1/2)sin(2v)] [0, π/2]
= 8 [(π/2) - (1/2)sin(π) - (0 - (1/2)sin(0))]
= 8 [(π/2) - (1/2)sin(π)]
= 8 [(π/2) - 0]
= 8 (π/2)
= 4π
Therefore, the surface area of the portion of the sphere described by the given parametric equations is 4π.