Question #cb563

1 Answer
Jun 10, 2017

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


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:


#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#


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)#.