Find the surface area of revolution about the x-axis of y=7sin(7x) over the interval 0 <= x <= pi/7

Answer 1

Has the problem been copied correctly?

The function f(x) = 7sin(7x) is just SCREAMING to have its antiderivative taken!

(The antiderivative is -cos(7x).)

But in an "area of surface of revolution" problem, we have to take the derivative, not the antiderivative.

(The derivative is 49cos(7x).)

The formula for the area of a surface revolution is

2π ∫ (from x=a to x=b) y √(1 + (dy/dx)^2) dx

In this problem, we have

y = f(x) = 7sin(7x)

dy/dx = 49cos(7x)

(dy/dx)² = 2401cos²(7x)

So the area of the surface of revolution is

2π ∫ (from x=0 to x=π/7) 7sin(7x) √(1 + 2401cos²(7x)) dx,

which would be unreasonably painful to evaluate.

(Its value is 2π ((√2402) + (arcsinh 49)/49), which is approximately 308.528)

Answer 2

surface area = integral of 7sin7x dx evaluated between x=0 and x=pi/7

= integral of sin7x 7dx from 0 to pi/7

let u =7x du=7dx, new u limits are 7(0) and 7(pi/7)=pi, or 0 and pi

= integral of sinu du from 0 to pi

= -cosu evaluated from 0 to pi

= -cospi - (-cos0)

= -(-1) - (-1)

= 1+1

= 2

Answer 3

Hi,

y=7sin(7x)

area = ∫7sin(7x) dx = -cos7x bounded from x=0 to x= Π/7

area = -cos 7(Π/7) - [ -cos7(0) ]

area = +1 +1 = 2

I hope it is useful,

Happy New Year