Respuesta :
Convert to polar coordinates using
[tex]x=r\cos\theta[/tex]
[tex]y=r\sin\theta[/tex]
Then the region [tex]R[/tex] is given by the set of points in polar coordinates
[tex]R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}[/tex]
The integral is then
[tex]\displaystyle\iint_R(x^2-2y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3(r^2\cos^2\theta-2r^2\sin^2\theta)r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\frac12\int_0^{\pi/2}\int_2^3(3\cos2\theta-1)r^3\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\frac12\left(\int_0^{\pi/2}(3\cos2\theta-1)\,\mathrm d\theta\right)\left(\int_2^3r^3\,\mathrm dr\right)[/tex]
[tex]=-\dfrac{65\pi}{16}[/tex]
To solve this integral we must change to polar coordinates, making:
x = r×cosθ and y = r× sinθ
The solution is:
∫∫R) ( x² - 2×y² ) ×dA = 59×π/8 square units
∫∫R) ( x² - 2×y² ) ×dA ⇒⇒ ∫∫R) ( x² - 2×y² ) ×dx×dy
If x = r×cosθ and y = r× sinθ dA = r×dr×dθ
And integrations limits: 2 ≤ r ≤ 3 0 ≤ θ ≤ π/2
r varies between 2 and 3 and θ between 0 and π/2
According to this:
∫∫R) ( x² - 2×y² ) ×dA = ∫₀(π/2) ∫₂³[ r²×cos²θ - 2×r²×sin²θ]×r×dr×dθ
∫∫R) ( x² - 2×y² ) ×dA = ∫₀(π/2) ∫₂³[ r³ × dr ( cos²θ - 2×sin²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ∫₀(π/2) (1/4)×r⁴ |₂³ ×( cos²θ - 2×sin²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 81/4 - 4 )× ∫₀(π/2) ( cos²θ - 2×sin²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 65/4) × ∫₀(π/2) ( 1 - 3×sin²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 65×π/8) - ∫₀(π/2) 3×sin²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 65×π/8) - 3× ∫₀(π/2) ( 1- cos²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 65×π/8) - (3×π)/2 +∫₀(π/2) cos²θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 65×π/8) - (3×π)/2 + 3 ×∫₀(π/2) ( 1/2 + 1/2)×cos2θ)×dθ
∫∫R) ( x² - 2×y² ) ×dA = ( 65×π/8) - (3×π)/2 + (3×π)/4 + (1/2) ×(1/2) ×0
∫∫R) ( x² - 2×y² ) ×dA = 59×π/8
Related Links: https://brainly.com/question/16728841
