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The application of calculus -- polar coordinates for area Why is the area of the graph enclosed by the curve r = 2acos @ (a > 0) 1 / 2 ∫ (2acos @) ^ 2D @ (the integral interval is - Π / 2 to Π / 2, not 0 to 2 Π?) thank you!
Because the graph of your curve is a circle that deviates from the origin
The radius is a and the center of the circle is (a, 0)
When your @ takes all the values from - Π / 2 to Π / 2, it just takes all the points on the circle
So the interval is - Π / 2 to Π / 2
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