Question

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?

Answer 1

The definite integral is `2int_3^5sqrt(25 - x^2)dx` .

Explanation:

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

We know that the equation for our circle is:
`x^2 + y^2 = 25`

This means that for any `x` value we can determine the two `y` values above and below that point on the x axis using:
`y^2 = 25 - x^2`
`y = sqrt(25-x^2)`

If we imagine that a line drawn from the top of the circle to the bottom with constant `x` value at any point, it will have a length of twice the `y` value given by the above equation.

`r = 2sqrt(25 - x^2)`

Since we are interested in the area between the line `x = 3` and the end of the circle at `x = 5`, those will be our integral boundaries. From that point on, writing the definite integral is simple:

`A = int_3^5rdx = 2int_3^5sqrt(25 - x^2)dx`

Answer 2

As an alternative, in polar

`=25int_{0}^{arcsin (4/5)} \ d psi - 12`

Explanation:

you can do it in polar too

the circle in polar is r = 5 and using the simplest formulation of area `A = 1/2 int \ r^2 (psi) \ d psi` becomes, using the symmetry about the x axis

`A = 2 times( 1/2 int_{0}^{arcsin (4/5)} \ 5^2 \ d psi - color{red}{1/2 * 3 * 4})`

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

`=25int_{0}^{arcsin (4/5)} \ d psi - 12`

`= 25 [psi]_{0}^{arcsin (4/5)} - 12`

`= 25 arcsin (4/5) - 12`