Final answer:
The surface area generated by rotating the curve x = 6t² and y = 4t³ about the y-axis is found by expressing x as a function of y and using the surface of revolution formula, integrating the expression with respect to y between the limits of 0 and 4(5)³.
Step-by-step explanation:
To find the surface area generated by rotating the curve x = 6t², y = 4t³ about the y-axis for 0 ≤ t ≤ 5, we can use the formula for the surface area of a surface of revolution. The formula for a surface rotating around the y-axis is given by:
S = 2π ∫ x ∙ √(1 + (dx/dy)²) dy
First, we need to express x as a function of y. Since y = 4t³, we can solve for t to get t = √[√(y/4)]. Substituting this into x = 6t², we have x = 6(√[√(y/4)])², which simplifies to x = 6y²/16.
To find dx/dy, we differentiate x with respect to y: dx/dy = ½(6/16)y⁻½.
Substituting x and dx/dy into the surface area formula, we get:
S = 2π ∫ (6y²/16) ∙ √(1 + (3/16)²y⁻’) dy, evaluated from y = 0 to y = 4(5)³ since t ranges from 0 to 5.
Now, we integrate this expression over the given limits to find the exact surface area. The integration involves using proper integral calculus techniques to evaluate the integral.