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

Question

Ezekiel Alexander · Accepted Answer

#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):

Isabella Ackerman · Answer

undefined The surface area produced by rotating $ f(x) = \sin(x) - \cos(x) $ around the x-axis over the interval $[0, \frac{\pi}{4}]$ is approximately $2.368$ square units.

Emma Aldridge · Answer

undefined The surface area produced by rotating $f(x) = \sin(x) - \cos(x)$ around the x-axis on the interval $[0, \frac{\pi}{4}]$ can be calculated using the formula for the surface area of a curve rotated around the x-axis, which is $\int_{a}^{b} 2\pi y \sqrt{1 + (y')^2} \, dx$, where $y = f(x)$ and $y' = \frac{dy}{dx}$.