Question

6. Use polar coordinates to evaluate: a) JJRdA/(x²+y²+1) where R is the region in the upper second quadrant between concentric circles (origin- centered) of radii 2 and 3. Sketch: 5 points b) JJR2x�

Answer 1

a) We can express the region R using polar coordinates with limits on θ between 0 and π/2, and limits on r between 2 and 3.

Hence,

we have:

`JJRdA/(x²+y²+1) = ∫[θ=0]^[π/2] ∫[r=2]^[3] (r / (r² + 1)) drd`

Using substitution method,

let u = r² + 1.

Then,

du = 2r dr.

We can then use this substitution in our integral to get:

JRdA/(x²+y²+1) = ∫[θ=0]^[π/2] \

∫[u=5]^[10] (1/2)

du dθ = ∫[θ=0]^[π/2] (1/2)

[u]5^10 dθ = ∫[θ=0]^[π/2]

(1/2) (10 - 5)

dθ= 5/4 (π/2)= (5π)/8b)

We can write the given function in polar form as:

2x² + y² = 2r² cos² θ + r² sin² θ

= r² (2 cos² θ + sin² θ) =

r² (2 cos² θ + 1 - cos² θ)

= r² (cos² θ + 1)

Hence,

the integral becomes:

JJR2x² + y²

dA = ∫[θ=0]^[2π] ∫[r=0]^[2 cos θ] r² (cos² θ + 1)

drdθ= ∫[θ=0]^[2π] [(2/3) cos³ θ + 2 cos θ]

dθ= (8π)/3

We can sketch the two regions as follows:1.

Region R lies in the upper second quadrant between the concentric circles (origin-centered) of radii 2 and 3.2.

Region R2 lies inside the circle x² + y² = 4.

To know more abo