also,
using Mathematica also, we get the same result. ( screenshot is attached below).
1. Evaluate x+1 using Mathematica. I. (4+1+y)d4; R DA; R = {(x,y): 0 5xs1, 2s y...
1. Evaluate S SR(5 – y)dA with R= {(x, y)|0 SX 55,0 Sy < 4} by identifying it as the volume of a solid and then calculating the volume geometrically.
Evaluate: vr y-x dA , y + 2x+1 where R is the parallelogram bounded by y-x-2, y-x-3, y + 2x = 0, andy+2x=4.
Evaluate: vr y-x dA , y + 2x+1 where R is the parallelogram bounded by y-x-2, y-x-3, y + 2x = 0, andy+2x=4.
Using Change of Variables..Evaluate ∫∫ R 15y/x dA where R is the region bounded by xy = 2, xy = 6 , y = 4 and y =10 usingthe transformation x=v , y=2u/3v.
1. Use polar coordinates to evaluate the double integral dA z2 +y where R is the region in the first quadrant bounded by the graphs x = 0, y = 1, y=4, and y V3z.
1. Use polar coordinates to evaluate the double integral dA z2 +y where R is the region in the first quadrant bounded by the graphs x = 0, y = 1, y=4, and y V3z.
Evaluate the integral using a change of variables. Z ZR (x + y) sin(x − y) dA (Z's are integrals) where R is the triangular region with vertices (−1, 1), (1, 1), and (0, 0).
QUESTION 4 Use the appropriate transformation to evaluate SX (2x + y)(x - y)dA where R is the region bounded by the line y = 4 - 2x, y = 7 - 2x, y = x - 2 and y = x +1. (8 marks)
3. (1.5 points) Evaluate the integral using a change of variables. (x + y)ez?-y dA JJR where R is the polygon with vertices (1,0), (0, 1), (-1,0), and (0, -1).
Let R be the region bounded by x + y=1, x - y=1, x+y=3, x-y=-1 evaluate the integral
s(x+ y)2sen2 (x - y)dA
s(x+ y)2sen2 (x - y)dA
Evaluate Double Integrals of sqrt(36 − x^2) dA, where R = [0, 6] × [−5, 4], using GEOMETRY only.
arctan log (?)) for any r > 1: Problem 4. Given that y = y(x) a) Evaluate r = *(y) at y = /4; b) Evaluate da/dy at y = 0.