What is the volume of the solid produced by revolving #f(x)=abssinx-abscosx, x in [pi/8,pi/3] #around the x-axis?

Question

1800 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T07:11:43

#pi/48( 10 pi-12(1+sqrt 2))=0.16005# cubic units, nearly.

I supplied the missing the factor 1.2 in the integral of sin 2x now. I

am sorry for missing it, in my earlier version..

#[pi/8, pi/3] in (0, pi/2)#.

Both sin x and cos x are > 0 in (0, #pi/2#). So,

#|sin x|-|cos x|=sin x - cos x, x in [pi/8, pi/3] #.

Now, the volume of solid of revolution is

#pi int (sinx -cos x)^2 d x#, between the limits #pi/8 and pi/3#

#=pi int (sin^2x+cos^2x-2 sinx cos x) d x#, between the limits

#=pi int (1-sin 2x) d x#, between the limits

#=pi [x+1/2cos 2x]#, between the limits

#=pi((pi/3-pi/8)+1/2(cos (2/3pi)-cos(pi/4))#

#=pi (5/24pi+1/2(-1/2-1/sqrt 2))#

#=pi/48(10pi-12(1+sqrt2))# cubic units.

Answer 2 · 2016-09-02T11:17:08

Interesting fact.

Comparing #f(x) = sinx-cosx# for #x in [pi/8,pi/3]# with
#y = y_1 + (y_2-y_1)/(x_2-x_1)(x-x_1)# in which

#x_1 = pi/8, y_1 = f(x_1) = sin(pi/8)-cos(pi/8)#
#x_2 = pi/3, y_2 =f(x_2) = sqrt(3)/2-1/2#

we observe a maximum deviation of #0.007# nearly

Attached a plot with #f(x)# and #y#

The difference between the computed volumes using either function is in the order of #0.0032# units

enter image source here