# How do you write the definite integral to find the smaller area cut from the circle #x^2 + y^2 = 25# by the line x = 3?

The definite integral is

There are always multiple ways to approach integration problems, but this is how I solved this one:

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As an alternative, in polar

you can do it in polar too

the circle in polar is r = 5 and using the simplest formulation of area

where the red bit is as shown shaded in red on the drawing

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To find the smaller area cut from the circle (x^2 + y^2 = 25) by the line (x = 3), you can write the definite integral as follows:

[ \int_{-4}^{3} \sqrt{25 - (x)^2} , dx ]

This integral represents the area between the curve of the circle and the line (x = 3), from (x = -4) to (x = 3).

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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.

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