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?

Jul 10, 2016

The definite integral is $2 {\int}_{3}^{5} \sqrt{25 - {x}^{2}} \mathrm{dx}$ .

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.

$r = 2 \sqrt{25 - {x}^{2}}$

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:

$A = {\int}_{3}^{5} r \mathrm{dx} = 2 {\int}_{3}^{5} \sqrt{25 - {x}^{2}} \mathrm{dx}$

Jul 10, 2016

As an alternative, in polar

$= 25 {\int}_{0}^{\arcsin \left(\frac{4}{5}\right)} \setminus d \psi - 12$

Explanation:

you can do it in polar too

the circle in polar is r = 5 and using the simplest formulation of area $A = \frac{1}{2} \int \setminus {r}^{2} \left(\psi\right) \setminus d \psi$ becomes, using the symmetry about the x axis

$A = 2 \times \left(\frac{1}{2} {\int}_{0}^{\arcsin \left(\frac{4}{5}\right)} \setminus {5}^{2} \setminus d \psi - \textcolor{red}{\frac{1}{2} \cdot 3 \cdot 4}\right)$

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

$= 25 {\int}_{0}^{\arcsin \left(\frac{4}{5}\right)} \setminus d \psi - 12$

$= 25 {\left[\psi\right]}_{0}^{\arcsin \left(\frac{4}{5}\right)} - 12$

$= 25 \arcsin \left(\frac{4}{5}\right) - 12$ 