Area of a surface of revolution

Question

Use Simpson's Rule with n = 10

 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare your answer with the value of the integral produced by a calculator or computer. (Round your answers to the nearest whole number.)

y =  x⁵/5,  0 ≤ x ≤ 5 

Simpson's Rule     ?

calculator approximation ?

I tried solving this question and I got 520.9 but because it says to round to the nearest integer I submitted 521 but my teacher said it's the wrong answer. I also don't know how to calculate the calculator approximation.

Answer 1

First make sure you're using the right formula for the surface area of a solid of revolution. If the function you're revolving is y=f(x), the form of the integral you're approximating would be:

SA= ∫ 2π•y*√ (1+(dy/dx)^2)dx, and the integral is done over the interval [a,b], where a is the lower limit of integration and b is the upper limit of integration.

To approximate the integral using Simpson's Rule, you'll the following formula:

S₁₀=(Δx/3)•[y(x₀)+4y(x₁)+2y(x₂)+4y(x₃)+2y(x₄)+4y(x₅)+2y(x₆)+4y(x₇)+2y(x₈)+4y(x₉)+y(x₁₀)]

where Δx=(b-a)/n, with n being the number of partitions.

As for how to do the calculator approximation, I assume they want you to plot the integrand of the above integral and then have the calculator calculate the area under the plot. On the Ti-84, graph the integrand, then push [2nd][trace]. Then scroll to option 7 under "Calculate".

Note that the answer you got is just the area under the curve, y=x⁵/5, not the surface area of the solid produced when that function is revolved around the x-axis. The answer you should get is approximately 2360 times greater.