Use a double integral to find the area of the region.
The region inside the circle
(x − 2)2 + y2 = 4
and outside the circle
x2 + y2 = 4
Is the equation to evaluate integral(rcos(theta)*rsin(theta)*rdrdthe… Thanks.
The region inside the circle
(x − 2)2 + y2 = 4
and outside the circle
x2 + y2 = 4
Is the equation to evaluate integral(rcos(theta)*rsin(theta)*rdrdthe… Thanks.
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The differential area is rdrdΘ. The limits of integration for Θ and r are determined by the two circles, their intersections and/or overlaps.
Look at the units of the integrand you propose. r*r*r*dr gives the integral units of Length^4. You know that cannot be correct since the units of area are Length^2.
Look at the units of the integrand you propose. r*r*r*dr gives the integral units of Length^4. You know that cannot be correct since the units of area are Length^2.
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I received the integral
integral [-π/3, π/3] (½ * ( 4 * cos Θ )² - 2² ) dΘ
after integration of the dr term.
integral [-π/3, π/3] (½ * ( 4 * cos Θ )² - 2² ) dΘ
after integration of the dr term.
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No