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?

2 Answers
Jul 10, 2016

Answer:

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#

Jul 10, 2016

Answer:

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#

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