Question

Question #cb563

Answer 1

I got about `17.25`, and Wolfram Alpha also does.

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DISCLAIMER: LONG ANSWER!

The surface area is given by:

`S = 2pi int_(a)^(b) f(x)sqrt(1 + ((dy)/(dx))^2)dx`

With `f(x) = sinx`, `f'(x) = cosx`, so

`S = 2pi int_(0)^(4) sinxsqrt(1 + cos^2x)dx`.

But since the graph passes through the `x` axis at `x = 0,pi` in `[0,4]`...

graph{y = sinx * sqrt(1 + (cosx)^2) [-0.396, 4.47, -1.087, 1.347]}

...we'll split this integral into two parts.

`S = S_1 + S_2`

`= 2pi int_(0)^(pi) sinxsqrt(1 + cos^2x)dx + 2pi int_(pi)^(4) sinxsqrt(1 + cos^2x)dx`

Let:

`u = cosx`
`du = -sinxdx`

This gives:

`S_i = -2pi int sqrt(1 + u^2)du`

One more substitution. This is of the form `sqrt(a^2 + u^2)`, so let:

`u = tantheta`
`du = sec^2thetad theta`
`sqrt(1 + u^2) = sqrt(1 + tan^2theta) = sectheta`

Therefore:

`S_i = -2pi int sec^3thetad theta`

If you do not know this integral, I have it here:

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`color(green)(intsec^3tdt)`

Let:

`u = sect`
`dv = sec^2tdt`
`du = sect tantdt`
`v = tant`

`= sect tant - intsect tan^2tdt`

`= sect tant - intsect(sec^2t - 1)dx`

`= sect tant - intsec^3tdt + int sectdt`

`2int sec^3tdt = sect tant + int sectdt`

`int sec^3tdt = 1/2(sect tant + ln|sect+tant|) + C`

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So, what we have is:

`S_i = -cancel(2)pi [1/cancel(2) (secthetatantheta + ln|sectheta + tantheta|)]`

Back-substitution gives:

`= -pi [sqrt(1 + u^2)cdotu + ln|sqrt(1 + u^2) + u|]`

`= -pi cosxsqrt(1 + cos^2x) - piln|sqrt(1 + cos^2x) + cosx|`

Evaluating from `x = 0` to `pi`:

`color(green)(S_1) = |[-pi cosxsqrt(1 + cos^2x) - piln|sqrt(1 + cos^2x) + cosx|]|_(0)^(pi)`

`= [-pi cospisqrt(1 + cos^2 pi) - piln|sqrt(1 + cos^2 pi) + cospi|] - [-pi cos0sqrt(1 + cos^2 0) - piln|sqrt(1 + cos^2 0) + cos0|]`

`= [pi sqrt(2) - piln|sqrt(2) - 1|] - [-pi sqrt(2) - piln|sqrt(2) + 1|]`

`= 2pi sqrt(2) - piln|sqrt(2) - 1| + piln|sqrt(2) + 1|`

`= color(green)(2pi sqrt(2) - piln|(sqrt(2) - 1)/(sqrt(2) + 1)|`

This is approximately `14.44`.

Finally, evaluating from `x=pi` to `x=4`:

`color(green)(S_2) = |[-pi cosxsqrt(1 + cos^2x) - piln|sqrt(1 + cos^2x) + cosx|]|_(pi)^(4)`

`= [-pi cos4sqrt(1 + cos^2 4) - piln|sqrt(1 + cos^2 4) + cos4|] - [-pi cospisqrt(1 + cos^2 pi) - piln|sqrt(1 + cos^2 pi) + cospi|]`

`= [-pi cos4sqrt(1 + cos^2 4) - piln|sqrt(1 + cos^2 4) + cos4|] - [pi sqrt(2) - piln|sqrt(2) - 1|]`

`= -pi cos4sqrt(1 + cos^2 4) - pi sqrt(2) - piln|sqrt(1 + cos^2 4) + cos4| + piln|sqrt(2) - 1|`

`= color(green)(-pi (cos4sqrt(1 + cos^2 4) + sqrt2) - pi ln|(sqrt(1 + cos^2 4) + cos4)/(sqrt(2) - 1)|)`

This value is around `-2.83`, but surface area is nonnegative, so we take the absolute value to get `2.83`.

Therefore, the total surface area should be:

`color(blue)(S) = 2pi sqrt(2) - piln|(sqrt(2) - 1)/(sqrt(2) + 1)| + pi (cos4sqrt(1 + cos^2 4) + sqrt2) + pi ln|(sqrt(1 + cos^2 4) + cos4)/(sqrt(2) - 1)|`

or about `color(blue)(17.25)`.