Question

How do you find the area of the surface generated by rotating the curve about the y-axis `y=cx+d, a<=x<=b`?

Answer 1

The surface area for `y = cx + d` rotated around the `y` axis for a given interval `x in [a,b]` is given by:

`S = pi(b^2 - a^2)sqrt(c^2 + 1)`, `" "" "c > 0`

(Note: If `c = 0` and `a = 0`, then the surface area reduces to the area of a circle of radius `b`.)

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If `x` is in `[a,b]` then assuming `c > 0`:

`y_(min) = ca + d`

`y_(max) = cb + d`

(If `c < 0`, then the revolution around the `y` axis would produce the same surface area by symmetry of `y = |c|x + d` compared to `y = -|c|x + d`. The value of `d` does not matter, since the rotation is around the `y` axis, and the shape will be the same regardless of the y-intercept.)

The surface area for a revolution around the `y` axis is

`S = 2 pi int_(ca + d)^(cb + d) f(y) sqrt(1 + ((dx)/(dy))^2)dy`

Evaluating the derivative, we get:

`((dx)/(dy))^2 = (d/(dy)[(y - d)/c])^2`

`= 1/c^2`

Therefore:

`S = 2 pi int_(ca + d)^(cb + d) (y - d)/c sqrt(1 + 1/c^2)dy`

`= (2 pi sqrt(1 + 1/c^2))/c int_(ca + d)^(cb + d) (y - d) dy`

`= (2 pi sqrt(1 + 1/c^2))/c |[y^2/2 - dcdoty]|_(ca + d)^(cb + d)`

`= (2 pi sqrt(1 + 1/c^2))/c [(cb + d)^2/2 - dcdot(cb + d)] - [(ca + d)^2/2 - dcdot(ca + d)]`

`= (2 pi sqrt(1 + 1/c^2))/c [(cb + d)^2/2 - dcdot(cb + d) - (ca + d)^2/2 + dcdot(ca + d)]`

`= (2 pi sqrt(1 + 1/c^2))/c [(b^2c^2 + 2bcd + d^2)/2 - bcd - d^2 - (a^2c^2 + 2acd + d^2)/2 + acd + d^2]`

`= (2 pi sqrt(1 + 1/c^2))/c [(b^2c^2 + 2bcd + cancel(d^2) - a^2c^2 - 2acd - cancel(d^2))/2 - bcd - cancel(d^2) + acd + cancel(d^2)]`

`= (2 pi sqrt(1 + 1/c^2))/cancel(c) [(cancel(c)(b^2c + 2d(b - a) - a^2c))/2 + cancel(c)d(a - b)]`

`= 2 pi sqrt(1 + 1/c^2) [(c(b^2 - a^2) + 2d(b - a))/2 + (2d(a - b))/2]`

`= 2 pi sqrt(1 + 1/c^2) [(c(b^2 - a^2))/2]`

`= 2 pi sqrt(c^2 + 1) cdot (b^2 - a^2)/2`

`= bb(color(blue)(pi (b^2 - a^2)sqrt(c^2 + 1)))`

For example, let's rotate `y = 2x + 3` in the interval `[x=0,x=2]` about the `y` axis using this formula and check our work.

`S stackrel(?" ")(=) pi (2^2 - 0^2)sqrt(2^2 + 1)`

`stackrel(?" ")(=) 4sqrt5pi`

In Wolfram Alpha, we get that it works!

Answer 2

`A = pi(b^2-a^2)sqrt(1+c^2)`

Explanation:

The function:

`y=cx+d`

represents a straight line. Rotating a straight line about `Oy` will form a cone, and so as we are rotating a closed interval `a le x le b` we will generate a frustum. The surface area of a frustum is easily calculated using the formula:

`A= pi(R+r)sqrt((R-r)^2+h^2)`

Where:

`R` is the lower radius
`r` is the upper radius
`h` is the vertical height

When:

`x=a => y=ca+d`
`x=b => y=cb+d`

Thus we have:

`r=a, R=b`
`h=(cb+d)-(ca+d) = cb-ca`

Thus, the SA, is:

`A = pi(a+b)sqrt((b-a)^2+(cb-ca)^2)`
`\ \ \ = pi(a+b)sqrt((b-a)^2+c^2(b-a)^2)`
`\ \ \ = pi(a+b)sqrt((b-a)^2(1+c^2))`
`\ \ \ = pi(a+b)(b-a)sqrt(1+c^2)`
`\ \ \ = pi(b^2-a^2)sqrt(1+c^2)`