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):
