√ √ 1. Use the transformation x = 2 u − p2/3 v, y = 2 u + p2/3 v to evaluate the integral RR 2 2 2 2 R(x − xy + y ) dA, where R is the region bounded by the ellipse x − xy + y = 2. √ √ ∂x ∂y 2 2 2 2 4 Jacobian of the transformation is ∂u ∂u = = √ + √ = √ . ∂x ∂y p p 3 3 3 ∂v ∂v − 2/3 2/3 The function x2 − xy + y2 in new variables becomes √ √ √ ( 2 u − p2/3 v)2 − ( 2 u − p2/3 v)(p2/3 v + p2/3 u) + ( 2 u + p2/3 v)2 = √ 2 2 √ 2 (2u2 − 4/ 3 uv + v2) − (2u2 − v2) + (2u2 + 4/ 3 uv + v2) = 2u2 + 2v2. 3 3 3 Therefore, the curve x2 − xy + y2 = 2 becomes 2u2 + 2v2 = 2, which is equivalent to u2 + v2 = 1. The integral is equal then to ZZ 4 (2u2 + 2v2) · √ du dv. u2+v2≤1 3 Passing to polar coordinates, we get that it is equal to

8 Z 2π Z 1 8 1 4π √ r2 · r dr dθ = √ · 2π · = √ . 3 0 0 3 4 3

2. Find a function f such that ∇f = F for F = y i + (x + z) j + y k. Use it to find the integral R t C F · dr along the curve C : r(t) = te i + (1 + t) j + t k, 0 ≤ t ≤ 1. ∂f ∂f ∂f We have to find a function f(x, y, z) such that ∂x = y, ∂y = x + z, ∂z = y. It follows from the first equality, that f is of the form xy + g(y, z). Substituting this into the second equality, we ∂g(y,z) get that x + ∂y = x + z. It follows that g(y, z) is of the form yz + h(z), hence f is of the form xy + yz + h(z). Substituting this into the third equation, we get that y + h0(z) is equal to y, which implies that h(z) is a constant. It follows that we can take f(x, y, z) = xy + yz = y(x + z). Beginning of the curve C is the point (0 · e0, 1 + 0, 0) = (0, 1, 0), its end is (e, 2, 1). It follows R that the integral C F · dr is equal to f(e, 2, 1) − f(0, 1, 0) = 2(e + 1) − 1(0 + 0) = 2(e + 1). 3. Find the area of the region bounded by the curve with parametric equation hx(t), y(t)i = 3 3 R hcos t i + sin t ji, 0 ≤ t ≤ 2π, using Green’s Theorem, by evaluating one of the integrals C −y dx, R 1 R C x dy, or 2 C (−y dx + x dy). Let us evaluate the third integral (one can also use the first two).

1 Z 2π (− sin3 t · (cos3 t)0 + cos3 t(sin3 t)0) dt = 2 0 1 Z 2π (3 sin3 t cos2 t sin t + 3 cos3 t sin2 t cos t) dt = 2 0 3 Z 2π (sin4 t cos2 t + cos4 t sin2 t) dt = 2 0 3 Z 2π 3 Z 2π sin2 t cos2 t(sin2 t + cos2 t) dt = sin2 t cos2 t dt = 2 0 2 0 3 Z 2π 3 Z 2π 1 − cos 4t 3π sin2 2t dt = dt = . 8 0 8 0 2 8 4. Find the area of the surface given by the parametric equation r(u, v) = uv i+(u+v) j+(u−v) k for u2 + v2 ≤ 1. 0 0 Let us find the area element dS = |ru × rv| du dv. We have

i j k 0 0 r × r = v 1 1 = h−2, u + v, v − ui. u v u 1 −1 √ Its length is p4 + (u + v)2 + (v − u)2 = 4 + 2u2 + 2v2. Hence, the area is equal to ZZ p4 + 2(u2 + v2) du dv. u2+v2≤1 Let us pass to polar coordinates: Z 2π Z 1 p Z 1 p 4 + 2r2 · r dr dθ = 2π 4 + 2r2 · r dr. 0 0 0 Make a substitution t = 4 + 2r2. Then dt = 4r dr, so that the area is equal to Z 6 √   6 2π π 2 3/2 π 3/2 3/2 t dt = t = (6 − 4 ). 4 4 2 3 t=4 3

RR 2 2 5. Evaluate S F · dS for F = −y j + z k, where S is the part of the paraboloid z = x + y below the plane z = 1 with upward orientation. The surface is the graph of the function g(x, y) = x2 + y2. It follows that dS = h−2x, −2y, 1i. The integral is equal to ZZ ZZ h0, −y, x2 + y2i · h−2x, −2y, 1i dx dy = 2y2 + x2 + y2 dx dy. x2+y2≤1 x2+y2≤1 Let us pass to polar coordinates:

Z 2π Z 1 Z 2π Z 1 (2r2 sin2 θ + r2)r dr dθ = (2 sin2 θ + 1)r3 dr dθ = 0 0 0 0 Z 1 Z 2π 1 Z 2π r3 dr (2 sin2 θ + 1) dθ = (1 − cos 2θ + 1) dθ = 0 0 4 0 1 Z 2π 1 (2 − cos 2θ) dθ = · 2 · 2π = π. 4 0 4

R 2 2 6. Use Stokes’ Theorem to evaluate C F · dr, where F = z i + y j + xy k, where C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 2) oriented counterclockwise as viewed from above. Curl of the vector field is

i j k

∂/∂x ∂/∂y ∂/∂z = hx, 2z − y, 0i.

z2 y2 xy Let us use the plane of the triangle as our surface. Equation of the plane is x + y + z/2 = 1. We can write it as z = 2 − 2x − 2y, and parametrize it by (x, y). Then dS = h2, 2, 1i dx dy. It follows that the integral is equal to ZZ ZZ ZZ hx, 2(2−2x−2y)−y, 0i·h2, 2, 1i dx dy = (2x+8−8x−8y−2y)dx dy = (8−6x−10y)dx dy, T T T where T is the projection of the triangle onto the xy-plane, i.e., the triangle in the xy plane with vertices (1, 0), (0, 1), and (0, 0). It follows that the integral is equal to

Z 1 Z 1−y Z 1 8 − 6x − 10y dx dy = (8(1 − y) − 3(1 − y)2 − 10y(1 − y)) dy = 0 0 0 Z 1 Z 1 7 12 4 (8 − 8y − 3 + 6y − 3y2 − 10y + 10y2) dy = (7y2 − 12y + 5) dy = − + 5 = . 0 0 3 2 3