220k views
0 votes
Find the area of the part of the surface r(u,v)=⟨8cosu,8sinu,v⟩ where 0≤u≤2π and 0≤v≤2. Give an exact answer. ________ sq. units

User Gen Wan
by
7.6k points

1 Answer

5 votes

Final answer:

To find the area of the surface, calculate the surface integral ∬_S ds, where S is the surface and ds is the element of surface area. Calculate the unit normal vector n(u,v) as the cross product of the partial derivatives of r with respect to u and v. Then, integrate the element of surface area over the given range to obtain the area A = 16π.

Step-by-step explanation:

To find the area of the given surface, we need to calculate the surface integral. The surface integral is given by ∬_S ds, where S is the surface and ds is the element of surface area. In this case, the surface is defined by the equation r(u,v) = ⟨8cosu, 8sinu, v⟩, with 0 ≤ u ≤ 2π and 0 ≤ v ≤ 2.

First, we need to find the unit normal vector to the surface, which can be calculated as:

n(u,v) = r_u × r_v

where r_u is the partial derivative of r with respect to u and r_v is the partial derivative of r with respect to v. After calculating the partial derivatives, we can obtain the unit normal vector:

n(u,v) = ⟨-8sinu, 8cosu, 0⟩

Next, we calculate the magnitude of n(u,v) to get the element of surface area:

ds = |n(u,v)| dudv = 8 du dv

Finally, we integrate ds over the given range of u and v:

A = ∬_S ds = ∫_{0}^{2π} ∫_{0}^{2} 8 du dv

Simplifying the integral gives us:

A = 16π

Learn more about Surface Area

User WinterChild
by
8.0k points

No related questions found