Question

Show transcribed data The double integral ∬ R​ xydxdy is defined as the larger of the two regions R above the x−axis bounded by the curves x² +y² =3,x² +y² =6,x+y=0.

(a) Using a transformation to polar coordinates the integral becomes of the form: a∫ 0b​ sin(cθ)dθ∫ de​ r r dr Specify the values of the following variables, corresponding to those highlighted in red above (enter answers in the boxes below) 1. a= 2. b= |3. c= 4. d= 5. e= 6. n=
(b) Hence evaluate the integral in (a) above and enter the answer in the box below. ∬ R xydxdy=

Answer 1

To evaluate the double integral ∬ R xy dxdy, we need to determine the limits of integration for the given region R. Using a transformation to polar coordinates, the integral becomes a∫ 0b sin(cθ)dθ ∫ de rdr. The values of a, b, c, d, e, and n are specified as follows: a = 0, b = π/4, c = 1, d = √3, e = √6, and n = 1. Finally, the integral can be evaluated using these values.

Step-by-step explanation:

To evaluate the double integral ∬ R xy dxdy, we first need to determine the limits of integration for the given region R. The region R is the larger area above the x-axis bounded by the curves x² + y² = 3, x² + y² = 6, and x + y = 0. To simplify the integral, we can use a transformation to polar coordinates. The integral in polar coordinates takes the form:

a∫0b sin(cθ)dθ ∫de rdr

Now, let's determine the values of the variables a, b, c, d, e, and n.

1. a = 0 (since the bounds of integration in the θ direction are from 0 to b)
2. b = π/4 (the angle at which the curves x² + y² = 3 and x + y = 0 intersect)
3. c = 1 (since we have one sine term in the integral)
4. d = √3 (the lower radial limit is the square root of the smallest radius of the region)
5. e = √6 (the upper radial limit is the square root of the largest radius of the region)
6. n = 1 (since there is no additional factor inside the integral)

Using these values, we can now evaluate the double integral:

∬ R xy dxdy = a∫0b sin(cθ)dθ ∫de rdr = 0∫π/4 sin(θ)dθ ∫√3√6 r r dr