# Plot the region R enclosed by y = cos(3x); y = cos(x); x = 0; x = pi Find all relevant intersection points. Find the volume of the solid S obtained by rotating the region R about the axis y = -1.

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### 1 Answer

You need to determine the inner and the outer radius such that:

inner radius: `r(x) = -1 - cos(3x)`

outer radius: `R(x) = -1 - cos x`

You need to evaluate the cross sectional area such that:

`A(x) = pi(R^2(x) - r^2(x))`

`A(x) = pi((-1 - cos x)^2 - (-1 - cos 3x)^2)`

`A(x) = pi(1 + 2cos x + cos^2 x - 1 - 2cos 3x - cos^2 3x)`

You need to evaluate the volume of solid of revolution such that:

`V(x) = int_0^(pi) A(x) dx`

Substituting `pi(1 + 2cos x + cos^2 x - 1 - 2cos 3x - cos^2 3x)` for A(x) yields:

`V(x) = int_0^(pi) pi(2cos x + cos^2 x - 2cos 3x - cos^2 3x) dx`

Using the linearity of integral yields:

`V(x) = pi( int_0^(pi) 2cos x dx + int_0^(pi) cos^2 x dx - int_0^(pi)2cos 3x dx- int_0^(pi)cos^2 3x dx)`

`V(x) = pi(2sin x + int_0^(pi) (1 + cos 2x)/2 dx - 2 (sin 3x)/3 - int_0^(pi) (1 + cos 6x)/2 dx)`

`V(x) = pi(2sin x + x/2 + (sin 2x)/4 - 2 (sin 3x)/3 - x/2 - (sin 6x)/12)|_0^pi`

`V(x) = pi(2sin x + (sin 2x)/4 - 2 (sin 3x)/3 - (sin 6x)/12)|_0^pi`

`V(x) = pi(2sinpi + (sin 2pi)/4 - 2 (sin 3pi)/3 - (sin 6pi)/12 - 2sin0- (sin 0)/4 +2 (sin 0)/3+ (sin 0)/12)`

Since `sin pi = sin 2pi = sin 3pi = sin 0 = 0` yields:

`V(x) = 0`

**Hence, evaluating the volume of the solid of revolution, under the given conditions yields `V(x) = 0` .**