Q. [-/1 Points]DETAILSMY NOTESSCALCET915.XP.3.019.Evaluate the iterated integral by converting to polar coordinates.∫01∫y2−y27(x+y)dxdy
Set Up Integral: First, we need to set up the integral in polar coordinates. The limits of integration for r will go from 0 to the circle defined by the equation x2+y2=2. The limits for θ will be from 0 to 2π because we're integrating over the first quadrant.
Convert to Polar: The equation x2+y2=2 in polar coordinates is r2=2, so r goes from 0 to 2. The integrand 7(x+y) in polar coordinates is 7(rcos(θ)+rsin(θ))=7r(cos(θ)+sin(θ)).
Write Integral: Now we can write the integral in polar coordinates: ∫0π/2∫027r(cos(θ)+sin(θ))drdθ.
Integrate with Respect to r: Let's integrate with respect to r first: ∫0π/2(∫027r(cos(θ)+sin(θ))dr)dθ.
Evaluate Integral for r: The integral of 7r(cos(θ)+sin(θ)) with respect to r is 27r2(cos(θ)+sin(θ)), evaluated from 0 to 2.
Integrate with Respect to theta: Plugging in the limits of integration for r, we get: 27(2)(cos(θ)+sin(θ))−0=7(cos(θ)+sin(θ)).
Evaluate Integral for θ: Now we integrate 7(cos(θ)+sin(θ)) with respect to θ from 0 to π/2:∫0π/27(cos(θ)+sin(θ))dθ.
Simplify Final Result: The integral of 7cos(θ) is 7sin(θ), and the integral of 7sin(θ) is −7cos(θ), both evaluated from 0 to π/2.
Simplify Final Result: The integral of 7cos(θ) is 7sin(θ), and the integral of 7sin(θ) is −7cos(θ), both evaluated from 0 to 2π. Plugging in the limits of integration for θ, we get:7sin(2π)−7sin(0)−(7cos(2π)−7cos(0)).
Simplify Final Result: The integral of 7cos(θ) is 7sin(θ), and the integral of 7sin(θ) is −7cos(θ), both evaluated from 0 to π/2. Plugging in the limits of integration for θ, we get: 7sin(π/2)−7sin(0)−(7cos(π/2)−7cos(0)). Simplifying, we have: 7(1)−7(0)−(7(0)−7(1))=7−0−(0−7)=7+7=14.
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