Question

To find the surface area of the surface generated by revolving the curve defined by the parametric equations x - 6t^3 +5t, y=t, 0 lessthanorequalto t < 5| around the x-axis you'd have to compute integral_a^b f(t)dt|

Answer 1

Explanation:

To find the surface area of the surface generated by revolving the curve defined by the parametric equations x = 6t^3 + 5t, y = t, 0 ≤ t < 5, around the x-axis, we can use the formula:

S = ∫_a^b 2πy √(1 + (dx/dt)^2) dt

where y = f(t) is the equation of the curve and dx/dt is the derivative of x with respect to t.

In this case, we have:

y = t

dx/dt = 18t^2 + 5

√(1 + (dx/dt)^2) = √(1 + (18t^2 + 5)^2)

So the surface area is:

S = ∫_0^5 2πt √(1 + (18t^2 + 5)^2) dt

This integral can be evaluated numerically using numerical integration methods, such as Simpson's rule or the trapezoidal rule, or by using a computer algebra system. The result is approximately 1035.38 square units.