What are the answers to these cal3 questions

1.) Evaluate the line integral, where C is the given curve. (Round your answer to one decimal place.)
∫c [xy^2] ds, C is the right half of the circle x2 + y2 = 16 oriented counterclockwise


2.) Evaluate the line integral, where C is the given curve.
∫c [xy dx + (x - y)]dy
C consists of line segments from (0, 0) to (2, 0) and from (2, 0) to (3, 2).


3.) Evaluate the line integral, where C is the given curve.
∫c [xyz]ds
C: x = 2sin(t), y = t, z = -2cos(t), 0 ≤ t ≤ π

1) Using x = 4 cos t, y = 4 sin t for t in [-π/2, π/2]:

Since ds = √((x')^2 + (y')^2) dt = √((-4 sin t)^2 + (4 cos t)^2) dt = 4 dt,

∫c xy^2 ds
= ∫(t = -π/2 to π/2) (4 cos t)(4 sin t)^2 * 4 dt
= 256 ∫(t = -π/2 to π/2) sin^2(t) cos t dt
= 256 * (1/3) sin^3(t) {for t = -π/2 to π/2}
= 512/3.

2) Do this in two parts:

i) (0, 0) to (2, 0) via x = t, y = 0 for t in [0, 2]:
∫ [xy dx + (x - y) dy]
= ∫(t = 0 to 2) [t * 0 * 1 + (t - 0) * 0] dt
= 0.

ii) (2, 0) to (3, 2) via y = 2x - 4 for x in [2, 3]:
∫ [xy dx + (x - y) dy]
= ∫(x = 2 to 3) [x(2x - 4) * 1 + (x - (2x - 4)) * 2] dx
= ∫(x = 2 to 3) (2x^2 - 6x + 8) dx
= (2x^3/3 - 3x^2 + 8x) {for x = 2 to 3}
= 17/3

So, ∫c [xy dx + (x - y) dy] = 0 + 17/3 = 17/3.

3) Since x = 2 cos t, y = t, z = -2 cos t for t in [0, π]:

ds = √((x')^2 + (y')^2 + (z')^2) dt = √((2 cos t)^2 + 1^2 + (2 sin t)^2) dt = √5 dt.

So, ∫c xyz ds
= ∫(t = 0 to π) (2 cos t) * t * (2 sin t) * √5 dt
= √5 ∫(t = 0 to π) t * 2 sin(2t) dt, via double angle identity
= √5 ∫(z = 0 to 2π) (z/2) sin z dz, letting z = 2t
= (√5/2) [-z cos z + sin z] {for z = 0 to 2π}
= 0.

I hope this helps!