Question

Compute the surface area of a surface generated by revolving the curve c(t) = (6 sinº(t), 6 cos?(t)) about the x-axis for 0<< (Use symbolic notation and fractions where needed.)

S=?

Answer 1

To compute the surface area of the surface generated by revolving the curve c(t) = (6 sin(t), 6 cos(t)) about the x-axis, we can use the formula for surface area of revolution. The surface area is given by S = 2π∫aby(t)√(1 + [y'(t)]2) dt

Step-by-step explanation:

To compute the surface area of the surface generated by revolving the curve c(t) = (6 sin(t), 6 cos(t)) about the x-axis, we can use the formula for surface area of revolution. The surface area is given by:

S = 2π∫aby(t)√(1 + [y'(t)]2) dt

First, we need to find the limits of integration by setting the y-coordinate equal to 0:

6 cos(t) = 0

cos(t) = 0

t = π/2, 3π/2

Now, we need to find the derivative of y(t):

y'(t) = d(6 cos(t))/dt = -6 sin(t)

Substituting these values into the formula, the surface area is:

S = 2π∫π/23π/2 6 cos(t)√(1 + (-6 sin(t))2) dt