Desperate help with integration!?

Question

I have to use the disc method to find the volume of the solid formed when the area enclosed by the curve y=sinx and the x-axis, between x=0 and x=pi , is rotated about the line y=2.

I just can't figure it out, although the area of the enclosed section is pi(4sinx - (sinx)2) if that helps at all.

thanks muchly

Answer 1

Volume of this figure rotated about the line y = 2 gives an 3D object whose radius about the line y = 2 is   (y1 – 2) at any instant of time thus the elemental volume at this point is given by   (pi) r^2  where r = (y1 – 2) thus the elemental volume would be

Integral {limits (0 to pi)} [(pi)  (y1 – 2) ^2 ]   dx  =  

                          

                 (Pi){ Integral {limits (0 to pi)} 4 dx + sin^2 x dx + 4sinx dx }



Which you can integrate to get [9/2(pi)^2 + 8(pi) ] which is the volume

Answer 2

hope this help

Answer 3

actually, you need to use the washer method.

V = π ∫ (outer radius)² - (inner radius)²

graph the equations and you'll see:

outer radius = 2

inner radius = 2 - sinx

so,

V = π ∫2² - (2 - sinx)² dx from 0 to π

V = π ∫4 - (4 - 4sinx + sin²x) dx

V = π ∫4 - 4 + 4sinx - sin²x dx

V = π ∫4sinx - sin²x dx

trig identity:

sin²x = 1/2 - (1/2)cos(2x)

V = π ∫4sinx - [1/2 - (1/2)cos(2x)] dx

V = π ∫4sinx - 1/2 + (1/2)sin(2x) dx

V = π [-4cosx - (1/2)x - (1/4)cos(2x)]

compute the limits and you'll get V ≈ 6.429π

Answer 4

its

int(pi(sin(x)-2)^2) from 0 to pi

its 19.2805

make it a good day