Evaluate the integral ∫∫r(x2−2y2)da, where r is the first quadrant region between the circles of radius 2 and radius 3. ∫∫r(x2−2y2)da

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: link

Convert to polar coordinates using

Then the region is given by the set of points in polar coordinates

The integral is then

it is not that hard all u have to do is plot the graph

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.