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SCALCET9 15.XP.3.019.
Evaluate the iterated integral by converting to polar coordinates.

int_(0)^(1)int_(y)^(sqrt(2-y^(2)))7(x+y)dxdy

[-/11 Points]\newlineDETAILS\newlineMY NOTES\newlineSCALCET99 1515.XP.33.019019.\newlineEvaluate the iterated integral by converting to polar coordinates.\newline01y2y27(x+y)dxdy \int_{0}^{1} \int_{y}^{\sqrt{2-y^{2}}} 7(x+y) d x d y

Full solution

Q. [-/11 Points]\newlineDETAILS\newlineMY NOTES\newlineSCALCET99 1515.XP.33.019019.\newlineEvaluate the iterated integral by converting to polar coordinates.\newline01y2y27(x+y)dxdy \int_{0}^{1} \int_{y}^{\sqrt{2-y^{2}}} 7(x+y) d x d y
  1. Set Up Integral: First, we need to set up the integral in polar coordinates. The limits of integration for rr will go from 00 to the circle defined by the equation x2+y2=2x^2 + y^2 = 2. The limits for θ\theta will be from 00 to π2\frac{\pi}{2} because we're integrating over the first quadrant.
  2. Convert to Polar: The equation x2+y2=2x^2 + y^2 = 2 in polar coordinates is r2=2r^2 = 2, so rr goes from 00 to 2\sqrt{2}. The integrand 7(x+y)7(x+y) in polar coordinates is 7(rcos(θ)+rsin(θ))=7r(cos(θ)+sin(θ))7(r\cos(\theta) + r\sin(\theta)) = 7r(\cos(\theta) + \sin(\theta)).
  3. Write Integral: Now we can write the integral in polar coordinates: 0π/2027r(cos(θ)+sin(θ))drdθ\int_{0}^{\pi/2}\int_{0}^{\sqrt{2}}7r(\cos(\theta) + \sin(\theta)) \, dr \, d\theta.
  4. Integrate with Respect to r: Let's integrate with respect to rr first: 0π/2(027r(cos(θ)+sin(θ))dr)dθ\int_{0}^{\pi/2}(\int_{0}^{\sqrt{2}}7r(\cos(\theta) + \sin(\theta)) \,dr) \,d\theta.
  5. Evaluate Integral for rr: The integral of 7r(cos(θ)+sin(θ))7r(\cos(\theta) + \sin(\theta)) with respect to rr is 72r2(cos(θ)+sin(θ))\frac{7}{2}r^2(\cos(\theta) + \sin(\theta)), evaluated from 00 to 2\sqrt{2}.
  6. Integrate with Respect to theta: Plugging in the limits of integration for rr, we get: 72(2)(cos(θ)+sin(θ))0=7(cos(θ)+sin(θ))\frac{7}{2}(2)(\cos(\theta) + \sin(\theta)) - 0 = 7(\cos(\theta) + \sin(\theta)).
  7. Evaluate Integral for θ\theta: Now we integrate 7(cos(θ)+sin(θ))7(\cos(\theta) + \sin(\theta)) with respect to θ\theta from 00 to π/2\pi/2:0π/27(cos(θ)+sin(θ))dθ.\int_{0}^{\pi/2}7(\cos(\theta) + \sin(\theta)) \, d\theta.
  8. Simplify Final Result: The integral of 7cos(θ)7\cos(\theta) is 7sin(θ)7\sin(\theta), and the integral of 7sin(θ)7\sin(\theta) is 7cos(θ)-7\cos(\theta), both evaluated from 00 to π/2\pi/2.
  9. Simplify Final Result: The integral of 7cos(θ)7\cos(\theta) is 7sin(θ)7\sin(\theta), and the integral of 7sin(θ)7\sin(\theta) is 7cos(θ)-7\cos(\theta), both evaluated from 00 to π2\frac{\pi}{2}. Plugging in the limits of integration for θ\theta, we get:\newline7sin(π2)7sin(0)(7cos(π2)7cos(0))7\sin(\frac{\pi}{2}) - 7\sin(0) - (7\cos(\frac{\pi}{2}) - 7\cos(0)).
  10. Simplify Final Result: The integral of 7cos(θ)7\cos(\theta) is 7sin(θ)7\sin(\theta), and the integral of 7sin(θ)7\sin(\theta) is 7cos(θ)-7\cos(\theta), both evaluated from 00 to π/2\pi/2. Plugging in the limits of integration for θ\theta, we get: 7sin(π/2)7sin(0)(7cos(π/2)7cos(0))7\sin(\pi/2) - 7\sin(0) - (7\cos(\pi/2) - 7\cos(0)). Simplifying, we have: 7(1)7(0)(7(0)7(1))=70(07)=7+7=147(1) - 7(0) - (7(0) - 7(1)) = 7 - 0 - (0 - 7) = 7 + 7 = 14.

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