SOLUTIONS TO PRACTICE EXAM 3

1. Find the mass of a flat solid sheet with density function δ(x, y) = 1+x kg/m^{2} bounded by the curves y = x^{2} and y = x + 2 (meters).

Graphing the region we obtain that the parabola is below the line and that they intersect when x^{2} = x + 2, or x = −1, 2 so that

M = ∫ ∫_{R} δ(x, y) dy dx = ∫_{−1}^{2} ∫_{x^{2}}^{x+2}(1 + x) dy dx

= ∫_{−1}^{2} y(1 + x)|_{x^{2}}^{x+2} dx = ∫_{−1}^{2} −x^{3} + 3x + 2 dx = 6.75.

2. Set up (but do not integrate) an integral representing the length of the arc with parametric equations

x = t,  y = t^{2},  z = t^{3},  1 ≤ t ≤ 3.

∫_{1}^{3} \sqrt{(dx/dt)^{2} + (dy/dt)^{2} + (dz/dt)^{2}} dt = ∫_{1}^{3} \sqrt{1 + 4t^{2} + 9t^{4}} dt.

3. Rewrite

∫_{x=0}^{6} ∫_{y=0}^{3−x/2} ∫_{z=0}^{2−x/3−2y/3} f(x, y, z) dz dy dx

as an iterated integral with innermost integration on y.

The region here is bounded by the three coordinate planes and the plane z = 2 − x/3 − 2y/3, that is x/6 + y/3 + z/2 = 1, so it has intercepts at x = 6, y = 3 and z = 2. So rewriting we get the integral

∫_{0}^{6} ∫_{0}^{2−x/3} ∫_{0}^{3−x/2−3z/2} f(x, y, z) dy dz dx.

4. An arrow is shot from the point P = (1, 1, 2) in the direction v⃗ = j⃗ − k⃗. At what point does the arrow strike the plane π given by the equation x − 2y + 3z = 10? Explain.

The tip of the arrow follows the line parametrized by r⃗ = ⟨1, 1, 2⟩ + t⟨0, 1, −1⟩, or x = 1, y = 1 + t and z = 2 − t for t ≥ 0. The arrow hits the plane when

10 = x − 2y + 3z = 1 − 2(1 + t) + 3(2 − t) = 5 − 5t,

so that t = −1. At that time the arrow would be at x = 1, y = 1 − 1 and z = 2 + 1, so that the arrow would strike the plane at (1,0,3). But we specified t ≥ 0, so that the arrow arrow will not hit the plane. Actually the arrow is shot in a direction away from the plane.

5. Sketch the vector field

F⃗ = \frac{y}{\sqrt{x^{2} + y^{2}}} i⃗ − \frac{x}{\sqrt{x^{2} + y^{2}}} j⃗.

This vector field will be ”circular” like 17.1.#13, except that it will go clockwise and the length of each vector will be

[(\frac{y}{\sqrt{x^{2} + y^{2}}} )^{2} + (\frac{x}{\sqrt{x^{2} + y^{2}}} )^{2}]^{1/2} = 1.

6. Compute the line integral ∫_{C} F⃗ · dr⃗ for F⃗ = xi⃗ − yj⃗ and C the parabola y = x^{2} − 1 from (0,-1) to (1,0).

Parametrize C by x = t, y = t^{2} − 1 and 0 ≤ t ≤ 1. So v⃗ = dr⃗/dt = 1i⃗ + 2tj⃗ and thus

∫_{C} F⃗ · dr⃗ = ∫_{0}^{1}[ti⃗ − (t^{2} − 1)j⃗] · [i⃗ + 2tj⃗] dt = ∫_{0}^{1} −2t^{3} + 3t dt = 1.