Let G be the region in the first octant that is inside both the cylinders x^2+y^2=1 and y^2+z^2=1. Use a triple integral to evaluate the volume of G.

Answer 1

The integral over x is then from 0 to 1,
the integral over y is from 0
to sqrt(1-x^2),
the integral over z is from 0 to
sqrt(1-y^2).

You integrate the volume element dxdydz.

Affter integrating over z, you are left with the integral over x and y of
sqrt(1-y^2) dydx.

Integrating over y then leaves you with the integral over x from 0 to 1 of
1/2 arcsin[sqrt(1-x^2)] + x/2 sqrt(1-x^2) dx

The integral of x/2 sqrt(1-x^2) is trivial, to compute the integral of the arcsin, you can substitute x = cos(t), that yields up to a factor an integral of t sin(t)dt, which can be computed using partial integration, or you can compute the integral of cos(p t) and then differentiate w.r.t. the parameter p and then set p equal to 1.

Answer 2

wow thank you! How did you find the limits for x and y?

Answer 3

You can see it by drawing a projection on the x-y plane. Then the region inside the cylinder parallel to the z-axis: x^2+y^2=1 is just what is inside the circle of radius 1 in the first quadrant (x and y positive).

The cylinder parallel to the x axis: y^2+z^2=1 is represented by the two lines y = 1 and y = -1. Everything inside it is in that cylinder. Since the entire quarter circle is inside this, this doesn't further contrain that projected region on the x-y plane you have to integrate over.

So, you can cover the region in the x-y plane by letting x run from 0 to 1 and y from 0 to sqrt(1-x^2). Then the limits on z are defined by the cylinder parallel to the x-axis, this is then from 0 to sqrt(1-y^2).