Evaluate the work integral F•dr Where F(x, y, z)= (3x +5y)i+(2x – 4y)j+ (x³ - z)k and C consists of the 3 line segments connecting the points (0,0,0) to (1,0,0) to (1,3,0) to (1, 3, 2)

Find (x2 + 4y)i + 3 y j) · dř where C consists of the three line segments from (1,0, 0) to (1, 4, 0) to (0, 4, 0) to (0, 4, 5). S(x? + 4y)i + 3y³}) · di =

Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables tou, v, w that puts q into the canonical form in `Sylvester's law of inertia'!. What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + 2), v = x+y, w = (x - y)/2 to find that q = u²+². Hence s = 2, t = 0 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q=u²v² w². Hence s = 1,t = 2 in Sylvester's law of intertia. O we let u = x, v= √2(x+y), w = x+y+z to find that q = u²v² + w²2. Hence s = 2, t = 1 in Sylvester's law of intertia. O None of the others apply O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.