Use a double integral to find the area of the region. The region inside the circle (x−1)2+y2=1 and outside the circle x2+y2=1

Question

tabuordg · Accepted Answer

Step1This equation is a circle shifted one unit to the right, according to a rudimentary grasp of mathematics. Because I realized that visualizing this in polar can be tough, I detailed each step for people who want to hand-jam.(x−1)2+y2=1(rcos⁡θ−1)2+r2sin2θ=1(r2cos2θ−2rcos⁡θ+1)+1r2sin2θ=1r2cos2θ+r2sin2θ=2rcos⁡θx2+y2=2rcos⁡θr2=2rcos⁡θr=2cos⁡θStep2Again, knowing that this is the equation of a circle with radius 1 is helpful, but if that is not clear, you can convert to polar.x2+y2=1r2=1r=1Step3Graphing this is simple and suggested for a better understanding of the integration limits. Make the necessary replacement and solve for theta to find your limits of integration for r.1=2cos⁡θcos⁡θ=12θ=−π3,π3Step 4Our Region can now be defined.R=(r,θ)∣1≤r≤2cos⁡θ,−π3≤θ≤π3Step 5Set up the integrals∫−π3π3∫12cos⁡θrdrdθStep 6Integrate with respect to r and run∫−π3π312r2∣12cos⁡θdθStep 7The power reduction formula is required at this step for further integration.∫−π3π32cos2θ−12dθStep 8You can double the integral and run from 0 to pi/3 using symmetry, but I&#039;ll show it without.∫−π3π32cos⁡2θ+12dθStep 9Use trig to find the values of the inputs12sin⁡2θ+12θ∣{−π3}π3Step10I went into greater detail with the problem setup than I did with the integration because I believe that is where the most difficulty arises. Please let me know if you encounter any issues with this solution.π3+32Result:&amp;nbsp;π3+32

Use a double integral to find the area of the

Answered question

Answer & Explanation

New Questions in High school geometry