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.
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.
wow thank you! How did you find the limits for x and y?
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).
To find the volume of the region G, we can set up a triple integral over the region.
First, let's visualize the region G. The cylinders x^2 + y^2 = 1 and y^2 + z^2 = 1 intersect in a circular region in the first octant. This circular region is the base of the volume we want to find.
To set up the triple integral, we need to determine the limits of integration for each variable. Let's start with z.
Since the region is bounded by the cylinder y^2 + z^2 = 1, the bounds for z will be from 0 to the height of the cylinder. The height of the cylinder is given by the maximum value of z when y = 1, which is sqrt(1 - 1^2) = 0.
Therefore, the bounds for z are from 0 to 0.
Next, let's determine the bounds for y. The region is bounded by the cylinder x^2 + y^2 = 1, which is a circle in the xy-plane centered at the origin with a radius of 1. Therefore, the bounds for y will be from 0 to the value of y on the circle, which is sqrt(1 - x^2).
Finally, we determine the bounds for x. Since the region G is in the first octant, x ranges from 0 to 1.
Putting it all together, the triple integral to evaluate the volume of G is:
∫∫∫ R 1 dz dy dx
where R is the region bounded by the cylinder x^2 + y^2 = 1 in the xy-plane, y ranges from 0 to sqrt(1 - x^2), and x ranges from 0 to 1.
Solving this triple integral will give us the volume of the region G.