What is the surface area produced by rotating $f(x)=sinx-cosx, x in [0,pi/4]$ around the x-axis?

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Answer 1 · 2018-06-02T15:59:05

$S_A=2piint_0^(pi/4)(sinx-cosx)*sqrt(2+sin2x)*dx=-3.7516$

The surface area due to $"x-axis"$ given by:

$color(red)[S_A=2piint_a^by*sqrt(1+(y')^2)*dx$

$y=sinx-cosx$

$y'=cosx+sinx$

$(y')^2=cos^2x+2sinx*cosx+sin^2x$

the interval of the integral $x in [0,pi/4]$

now let setup the interval of the definite integral to determine the surface area:

$S_A=2piint_0^(pi/4)(sinx-cosx)*sqrt(1+(cosx+sinx)^2)*dx$

$S_A=2piint_0^(pi/4)(sinx-cosx)*sqrt(1+cos^2x+2sinx*cosx+sin^2x)*dx$

$S_A=2piint_0^(pi/4)(sinx-cosx)*sqrt(1+1+sin2x)*dx$

$S_A=2piint_0^(pi/4)(sinx-cosx)*sqrt(2+sin2x)*dx$

$=2pi[(root(4)17*sin(arctan(1/4)/2))/2^(7/2)+(root(4)17*cos(arctan(1/4)/2))/2^(7/2)-sqrt(5)/8-sqrt(3)/sqrt(2)+1/sqrt(2)]$

$=2pi[(8root(4)17*sin(arctan(1/4)/2)+8root(4)17*cos(arctan(1/4)/2)-2^(7/2)*sqrt(5)-64sqrt(3)+64)/2^(13/2)]$

$=-3.7516$

show below the surface area revolving (shaded):

enter image source here

What is the surface area produced by rotating f(x)=sinx-cosx, x in [0,pi/4]f(x)=sinx-cosx, x in [0,pi/4] around the x-axis?