10. Verify Stokes' Theorem for: (a) F (2r2y) + (5ry+6)j + (z² + 2r2-3) and S is the surface of the paraboloid z = above the ry plane. 9-(²+ y²) (b) = (x+3y) + (2y + 4x); + 32³k and S is the hemispherical surface z = (25 - r² - y²).

4. Identify the quadric surface 2x² + 3y² = 4x + 2z². 2x²+36² 2x² - 4x +38² 2(x²-2y+1 A. Hyperboloid of two sheets B. Elliptic paraboloid C. Hyperbolic paraboloid D. Hyperboloid of one sheet E. Elliptic cone 2x²-4x72

1. Find the traces of the given surface in the planes x = k, y = k, z = k. Then identify the surface and sketch it. (a) x = y² + z² (b) x² - y²+z² = 1 (c) 9x2 y2 z² = 9 - (d) 4z²x² - y² = 1

Identify the quadric surface satisfying each of the given equations and then make a rough sketch to show its position and orientation. (a) y – x² + z² = 3 | (b) 9x2 + y? + 422 – 18x + 2y + 16z = 10 (c) z² = 4x² + y² + 8x – 2y + 4z -

Consider the quadric surface described by 2x2 + 6xy + 2y² + 3z² = 1. You are given that in canonical coordinates, the same surface is described by 5² + 3v2 – w? = 1. The principal axes are in the directions of the vectors B1, B2 and B3. What are B1, B2 and B3? None of these is correct. () B1 = 0, B2 = , B3 = () B1 B2 Вз (3) 1 B1 = 31 B2 = B3 = 1 -1 ()- 1 B1 = 3 1 ,B2 = -1},B3 = | 0 0.

4. Let F (0, 0, 2) and S be the part of the paraboloid z = x² + y² between the plan z = 1 and z = 4, oriented with the normal pointing away from the surface. Evaluate f fF.ds. (a) 0. (b) -T. -2/. (d) -67. (e) -15.

4. Consider the surface defined by the portion of the cone z = -Vx2 + y? that lies below the rectangle (x, y) E [–1, 1] x [2, 4]. (a) Find parametric equations for this surface. (b) Compute the area of this surface.

! Let (P) be the quadric surface given by the following equation y=5+x²+² Then: (P) is a circular paraboloid along y-axis This option (P) is an elliptic paraboloid along y-axis This option (P) is an elliptic paraboloid along z-axis

4. Find parametric equations of the line tangent to the surface z = y² + x³y at the point (2,1,9) whose projection on the xy-plane is parallel to the x-axix.

Given z=12+73x² is a curve in the xz-plane. If the equation of the surface generated by revolving the curve about the z-axis is a₁x²+₂y²+³²²+x+ay+z+a6=0, then 1+ +Σ₂ i=1 (Notice that you have to write the surface equation where the coefficient of z is one.) aj=