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M ASSACHUSETTS I NSTITUTE OF T ECHNOLOGY Interphase Calculus III Worksheet Instructor: Samuel S. Watson 31 July 2015 Topics: Surface integrals, change of variables The surface area of the graph of a function f ( x, y) above a region R is given by s 2 2 ZZ ∂f ∂f 1+ ( x, y) + ( x, y) dx dy ∂x ∂y R Explain this formula. 1. Find the surface area of the part of the surface z = x2 + 2y that lies above the triangular region T in the xy-plane with vertices (0, 0), (1, 0), and (1, 1). 2. Find the area of the part of the paraboloid z = x2 + y2 that lies under the plane z = 9. 3. Find the area of the part plane 3x + 2y + z = 6 that lies in the first octant. Changing coordinates 4. Suppose that we define functions u( x, y) and v( x, y) on R2 and identify the locations of points by their u and v values instead of their x and y values. Then ∂x ZZ ZZ ∂u f ( x, y) dx dy = f (u, v) det R R ∂y ∂u ∂x ZZ ∂v du dv = det f ( u, v ) R ∂y ∂v ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y −1 du dv. Explain this formula. 2 2 5. Use the change of variables x = u − v , y = 2uv to evaluate the integral 2 2 ZZ R y dA, where R is the region above by the x-axis bounded by the parabolas y = 4 − 4x and y = 4 + 4x. x−y over the square whose vertices are the four points of intersection x+y+z between the axes and the unit circle. 6. Find the integral of 7. Find ZZ R x2 dA, where R is the regino bounded by the ellipse 9x2 + 4y2 = 36, using the trans- formation x = 2u, y = 3v. 8. (Challenge problem) Let f be continuous on [0, 1], and let R be the triangular region with vertices (0, 0), (1, 0), (0, 1). Show that ZZ R f ( x + y) dA = Z 1 0 u f (u) du.