Va2 y da dy The region A is bounded by the curve: 2+y=Va 3. Evaluate C 2102 dz dy dz 4. Evaluate The solid V bounded by surfaces: z = 1-2, z = y , y = 0
Va2 y da dy The region A is bounded by the curve: 2+y=Va 3. Evaluate C 2102 dz dy dz 4. Evaluate The solid V bounded by surfaces: z = 1-2, z = y , y = 0
Solve the initial value problem dy + бу da 0, y(In 4-4.
Answer is E
7. Find the solution to the initial value problem dy da 6ry2(3ar2 + 2xy + 2y) 0 y(1) 3 A. 6ry2y2x = 37 B. ry y2 +x = 22 C. 3r2y2+ x3 + 2r2 + 2y = 21 D. y2ry y2 + x = 31 E. 3x2yxy2 y? = 27
please respond with explanations for each step. thank
you
Problem 4 Evaluate the line integrals (a) (10 points) y da 2ax dy, where C is the curve r(t) (2t + 1) i+ 3t2 j, 0t 1. (b) (10 points) (ryz) ds, where C is the line segment from the point (2, 1,0) to the point (4,3,6) (c) (10 points) F.dr,where F is the vector field F(x, y) = yi - rj and C is the curve given by r(t) t2i+...
. (5pont)Thedale integraltegralsovertherduis an improper integ da dy is an improper integral that could be defined as the limit of double integrals over the rectangle [0,t] x [0, t] as t-1. But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that Tl 2. (5 points) Leonhard Euler was able to find the exact sum of the series in the previous problem. In 1736 he proved that...
7. Assume x and y are functions of t. Evaluate dy/dt for each of the following. (a) y2 - 8x3 = -55, 5= -4,2 = 2, y = 3 (b) = 2, r = 4, y = 2 (e) cell = 2 - In 2 + In 2, 6,2 = 2, y = 0
Write an iterated integral to evaluate the integral || 2?;} dA where R is the triangular region with vertices (18,0), (13,9) and (18,13). R Select all that apply -7 5 162 5 162 18 9 5 2+ - [ ] 22,3 dA= 5 22 43 dy do -7 + R 5 5 S] =2,3 dA – S139 -7 + 5 5 162 -2 + 5 5 22y3 dy do R 13 22,3 dA= -7 5 22 162 5 dy do...
d1 dy 2. Solve the system dt dt2 da dt =t = 2 dy (25pts) + 3x + + 3y = dt
16. o integrad [**** The triple da dy dz describes the solid pictured at right. Rewrite as an equivalent triple integral in the following orders (DO NOT EVALUATE): 31 (a) dy dz dx (b) du dz dy 2. 16-2 21. Given dy da, 16- (a) Sketch the region of integration and write an equivalent iterated integral in the order dx dy. (You do not need to evaluate it!) (b) Now write it as an equivalent iterated integral in polar coordinates....
1. A -f(A),y-g(A), 0 < λ < 1]. S he tangent vector (dx/dA, dy/dA) does actually lie tangent to the curve. how that t curve is denned by 1*
1. A -f(A),y-g(A), 0