Using integrals, find the area of the circle #x^2 + y^2 = 1# ?
For symmetry reasons:
substitute:
As in the interval the cosine is positive:
So:
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To find the area of the circle (x^2 + y^2 = 1), you can use the integral method. Since the circle is symmetric about both the x-axis and the y-axis, you can focus on finding the area of one quarter of the circle (for example, the region where (x \geq 0) and (y \geq 0)) and then multiply by 4.
The equation of the circle suggests that (y = \sqrt{1 - x^2}) for (x) between -1 and 1.
So, the area (A) of one quarter of the circle can be found by integrating (y = \sqrt{1 - x^2}) with respect to (x) from 0 to 1.
[ A = \int_{0}^{1} \sqrt{1 - x^2} , dx ]
After integrating this expression, multiply the result by 4 to find the total area of the circle.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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