Evaluating Integral (S)

Question

Evaluate double integral_S 2y dS where S is the portion of y^(2)+z^(2) = 4 between x = 0 and x = 3-z.

Having problems with setup and then following through.

Answer 1

The surface y² + z² = 2² is a cylinder of radius 2 whose axis is the x axis. (Remember that in the plain old x-y plane, the equation x² + y² = radius² is a circle.) At the x = 3-z end of the cylinder, the cylinder is cut diagonally like one of those straws that you thrust through the hole in the plastic top of a paper cup or juice carton.

Half of this cylindrical surface has lies above the xz plane (i.e., has positive y values), and half lies below the xz plane (i.e., has negative y values). We therefore expect the value of the integral ∫∫_S 2y dS to be exactly zero.

Each point on the cylindrical surface has thee coördinates (x, y, z). (In fact, every point in three dimensional space has three (x, y, z) coördinates.) But a simpler way to describe the position of each point on the cylindrical surface is with two (x, θ) coördinates. Note that for any point on the cylindrical surface of radius 2, z=2sin θ and y=2cos θ. Therefore if we are given the two (x, θ) coördinates for any point on the cylindrical surface, we can reconstruct all three (x, y, z) coördinates.

(Remember that in the x-y plane, each point in a circular disk has two (x, y) coördinates. But a simpler way to describe the location of a point within a circular disk is with (r, θ) coördinates. Note that for each point in our circular disk, x=r sin θ and y=r cos θ. Therefore if we are given the two (r, θ) coördinates for any point in the disk, we can reconstruct the two (x, y) coördinates. And when we integrate across this circular disk, an area element dA = dr r dθ.)

On our cylindrical surface, the θ coördinate runs from θ=0 to θ=2π, because we want to go all the way around the cylinder. The x coördinate runs from x=0 (at the flat end of the cylinder) to x=3-z (at the pointy, sharpened end of the cylinder).

For example, when z=-2 (i.e., when θ = 3π/2, or in English, θ = 270 degrees), the x coördinate runs all the way from x=0 to to x=3-z=5, and the point (x=5 y=0, z=-2) is at the pointy tip of the cylinder, the first point on the leading edge of the cylinder to penetrate the hypothetical plastic top of the paper cup.

When z=2 (i.e., when θ = π/2, or in English, θ = 90 degrees), the x coördinate runs only from x=0 to to x=3-z=1, and the point (x=1, y=0, z=2) is the last point on the leading edge of the cylinder to penetrate the hypothetical plastic top.

Finally, when z=0 (i.e., when θ = 0 or π, or in English, θ = 0 degrees or 180 degrees), the x coördinate runs from x=0 to x=3-z=3.

We were told that the x coördinate runs from x=0 to x=3-z, but we can restate this by saying that the x coördinate runs from x=0 to x=3-z=3-2sin θ. (After all, as we saw earlier, z=2sin θ).

And the integrand 2y = 4 cos θ, because as we saw earlier, y=2cos θ.

And the radius of the cylinder is 2. (That's the 2 just before the dθ we're about to see.)

So the double integral we want is

∫ (from θ=0 to θ=2π) ( ∫ (from x=0 to 3-2sin θ)4cos θ dx) 2 dθ

In the inner integral ∫ (from x=0 to 3-2sin θ) 4cos θ dx, the θ is a constant with respect to x

Therefore 4cos θ is also a constant with respect to x in the inner integral, so we can take the 4cos θ out of the integral and move it to the front:

∫ (from x=0 to 3-2sin θ) 4cos θ dx

= (4cos θ) ∫ (from x=0 to 3-2sin θ) 1 dx

= (4cos θ)(3-2sin θ)

= 12cos θ - 8(sin θ)(cos θ)

Now let's do the outer integral.

Remember the trig identity sin (x+y) = (sin x)(cos y) + (cos x)(sin y)

Therefore sin (2θ) = sin (θ+θ) = (sin θ)(cos θ) + (cos θ)(sin θ)

Therefore sin (2θ) = 2(sin θ)(cos θ)

Therefore (sin θ)(cos θ) = (sin (2θ)) / 2

And the derivative of -(cos 2θ)/4 is (sin 2θ)/2.

So the derivative of (cos 2θ)/4 is -(sin 2θ)/2.

The outer integral is

∫ (from θ=0 to θ=2π) (12cos θ - 8(sin θ)(cos θ))2 dθ

We can already see that the value of this definite integral will be zero, since we're integrating from θ=0 to θ=2π, and the trig functions have the same value at 0 and 2π. But let's compute the definite integral anyway.

= 8 ∫ (from θ=0 to θ=2π) (3cos θ - 2(sin θ)(cos θ)) dθ

= 8 ∫ (from θ=0 to θ=2π) (3cos θ - (sin 2θ)) dθ

= 8(3sin θ + (cos (2θ)/2) (from θ=0 to θ=2π))

= 8(1/2 - 1/2)

= 8·0

= 0

which is the big zero we expected. Thank you for giving me an opportunity to review all of this.

Answer 2

This first thing I do when I look at double integrals is visualize the shape. Can you imagine (or construct out of a paper towel tube) the shape you get when you cut y²+z²=4 by the two planes you list?