another integral by volumes?

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=0, y= cos(6 x), x = {pi}/{12}, x = 0 about the axis y= -1

is this a washer problem?

Answer 1

Yes, it is a washer problem.

The inner radius is 1, the distance between y=0 and y=-1.

The outer radius is cos(6x) + 1, the distance between y=cos(6x) and y=-1.

The limits are 0 and pi/12.

The integrand is pi*[R(x)^2 - r(x)^2], where R(x) is the outer radius and r(x) is the inner radius.

For integrating cos^2(6x), rewrite it as (1+cos(12x))/2 first.

Answer 2

you're integrating z over the triangular section. comedian strip the triangle. word that y=2-x and y=x intersect on an analogous time as x=a million. you will see that interior the time of this section, y runs from x on the shrink line to 2-x on the precise line. Then x runs from 0 to a million. we've int (0,a million) int (x,2-x) of x^2+y^2 dydx inner int =x^2y + y^3/3 from x to 2-x =x^2(2-x)+(2-x)^3/3 - x^3 - x^3/3 =2x^2 + (2-x)^3/3 - 7x^3/3 Integrating this with comprehend to x we've 2x^3/3 - (2-x)^4/12 -7x^4/12 Then from 0 to a million we've 2/3 - a million/12 - 7/12 +sixteen/12 = 4/3

Answer 3

dV = π((1 + cos6x)^2 - 1)dx =
π(2cos6x + (1 + cos12x) / 2)dx =
V = π((sin6x)/3 + x + (sin12x/24) | x = 0 to x = π/12 =
π(1/3 + π/12 + 0)
V = 1.86966

It is a washer.