Calculus, Applications of Integration

Question

How would you find the volume of the solid of revolution formed when the curve 𝑦=(√sin𝑥) between 𝑥=0 and 𝑥=𝜋 is revolved about the 𝑥-axis?

Answer 1

The method to go with for this problem is under the concept of the Disk Integration Method about the x-axis. The principal idea behind it is if you really zoom in on the curve being rotated about the x-axis, the curve would look approximately linear. The "rectangular" approximation of integrating an average function is taken to be the same idea that as you zoom in, the curve will look linear-like and if you sum the areas of the rectangles as the number of areas approaches infinity then you will achieve the integral of that function and find the area under the curve to the x-axis. A similar yet different approach is taken in the Disk Integration Method (DIM). Swap the infinitesimal rectangles for infinitesimal cylinders with radius R(x) and thickness/height dx.

And the sum of the cylinders as the number of cylinders approaches infinity will give you the volume bounded as you rotate the y(x) curve about the x-axis.

So, the formula is as follows:

V = π * ∫_a^b [R(x)]^2 * dx

So:

R(x) = √[ sin( x ) ] ∴[R(x)]² = sin(x) ∴[R(x)]² dx = sin(x)*dx

∫₀^π [R(x)]^2 * dx = ∫₀^π sin(x) * dx = - cos( x )]₀^π = - [cos( π ) - cos( 0 )] = - [ -1 - (1) ] = -( -2 )

∴∫₀^π [R(x)]^2 * dx = 2

V = π * ∫₀^π [R(x)]^2 * dx = π * ( 2 ) ∴ V = 2 * π units³

I hope this helps! Message me in the comment section below if you have any questions!