Exercise 4. Implicit differentiation (15 pts) Given z - xy + yz + y = 2...
Exercise 4. Implicit differentiation (15 pts) Given z3 – xy + yz + y3 = 2 and z is a differentiable function in x and y. Then at (1,1,1) is: az дх a. 0 b. 1 1 с. 2 d. e. None of the above a. b. e.
Exercise 6. Double integral in rectangular coordinates (10 pts+10 pts) Let I = SMS secx dydx. 1) The region of integration of I is represented by the blue region in Oь. d
Exercise 6. Double integral in rectangular coordinates (10 pts+10 pts) Let I = S secx dydx. 1) The region of integration of I is represented by the blue region in: * Oь. C. O d.
1) The region of integration of I is represented by the blue region in: O a Oc. Od 2) By reversing the order of integration of I, we get: a 1 = $secx dxdy b. I = 8 secx dxdy c. 1 = secx dxdy d. 1 - IL secx dxdy Exercise 6. Double Integral in rectangular coordinates (10 pts 10 pts) Let I= secx dydx. 1) The region of integration of I is represented by the blue region in:...
Exercise 6. Double integral in rectangular coordinates (10 pts+10 pts) Let I secx dydx. 2) By reversing the order of integration of I, we get: a. I = $ S secx dxdy b. 1= SS secx dxdy c. IESU secx dxdy d. 1 = secx dxdy
Exercise 6. Double integral in rectangular coordinates (10 pts+10 pts) Let I = secx dydx. 2) By reversing the order of integration of I, we get: a. I = secx dxdy b. I = ('secx dxdy c. INSS secx dxdy d. I = So, secx dxdy
Exercise 6. Double integral in rectangular coordinates (10 pts+10 pts) Let I = SL, secx dydx. 2)By reversing the order of integration of I, we get: a. I = 16 secx dxdy b. I = foto secx dxdy c. 1 = 1secx dxdy d. 1 = SS, SS,' secx dxdy C. O d.
QUESTION 7 Using implicit differentiation to find az ax for x2 + xy -sin(z)-0 y + COSZ-X Х COSZ-X y COS2-X y +Z sinz-X
Let ∭E (yz)dV, where E = {(x,y,z)/ x = 1 - y^2 - z^2, x>=0} a. Sketch E, the solid of integration. b. Sketch D, the region of integration in the plane the solid is projected onto. c. Evaluate the integral using cylindrical coordinates.
Let f(x,y,z) = xy + z-5,x=r +2s, y = 2r - sec(s), z = s Then I is: ar a. r - sec(s) b. sec(s) c. r+s+sec(s) d. 4r + 4s - sec(s) a. b. C. Given zº – xy + y2 + y2 = 2 and z is a differentiable function in x and y. Then at (1,1,1) is: дх a. 0 b. 1 c. d. e. None of the above o a. o b. ♡ C. o d.