Finding Area by an Iterated Integral

[_N3WTON_]

Homework Statement

Use an iterated integral to find the area of the region bounded by the graphs of:
$f(x) = sin(x)$
$g(x) = cos(x)$
between:
$x = \frac{\pi}{4} and x = \frac{5\pi}{4}$
Find two solutions, one using a vertical representative rectangle and another using a horizontal representative rectangle.

Homework Equations

Definite Integral Equation

The Attempt at a Solution

I had no issue finding the solution using the vertical representative rectangle, here is my solution:
$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \int_{cos(x)}^{sin(x)} dy dx$
$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (sin(x) - cos(x)) dx$
$A = 2\sqrt{2}$
The problem I am having is trying to solve this problem using a horizontal representative rectangle. I understand that the solution will be the same, but I can't even seem to set up the problem properly. Would I have to define the two fractions:
$\frac{\pi}{4} and \frac{5\pi}{4}$
as functions of y and then make the trigonometric functions constant and then integrate?

 

[_N3WTON_]

Another idea I had was to make the two trigonometric functions functions of 'y' instead of 'x', but I do not know if that is the right approach...

 

[BvU]

Re problem statement: Do you mean between $x = \frac{\pi}{4}$ and ${\bf x} = \frac{5\pi}{4}$ ? I really can't see y = 5/4 $\pi$ as a sensible bound...

Re horizontal rectangles: If I make a drawing (you should do so too!) and give it a quarter turn, it looks as if I need to split up in three pieces: one from y = 1 to $\cos{\pi \over 4}$, one from there to $\cos{5\pi \over 4}$ and one from there to -1. bounds on x are then $\arcsin y$ and $\pi -\arcsin y$ for the first, etc.

 

[_N3WTON_]

Re problem statement: Do you mean between $x = \frac{\pi}{4}$ and ${\bf x} = \frac{5\pi}{4}$ ? I really can't see y = 5/4 $\pi$ as a sensible bound...

Re horizontal rectangles: If I make a drawing (you should do so too!) and give it a quarter turn, it looks as if I need to split up in three pieces: one from y = 1 to $\cos{\pi \over 4}$, one from there to $\cos{5\pi \over 4}$ and one from there to -1. bounds on x are then $\arcsin y$ and $\pi -\arcsin y$ for the first, etc.

wow I'm sorry, I messed up entering the equations, it should be
$x = \frac{5\pi}{4}$

 

[_N3WTON_]

Re problem statement: Do you mean between $x = \frac{\pi}{4}$ and ${\bf x} = \frac{5\pi}{4}$ ? I really can't see y = 5/4 $\pi$ as a sensible bound...

Re horizontal rectangles: If I make a drawing (you should do so too!) and give it a quarter turn, it looks as if I need to split up in three pieces: one from y = 1 to $\cos{\pi \over 4}$, one from there to $\cos{5\pi \over 4}$ and one from there to -1. bounds on x are then $\arcsin y$ and $\pi -\arcsin y$ for the first, etc.

ok, so the integral would end up looking something like:
$A = \int_{\cos\frac{pi}{4}}^{\cos\frac{5\pi}{4}} \int_{arcsin(y)}^{{\pi}-arcsin(y)} dx dy$
or would I need to make 3 separate integrals?

 

[LCKurtz]

ok, so the integral would end up looking something like:
$A = \int_{\cos\frac{pi}{4}}^{\cos\frac{5\pi}{4}} \int_{arcsin(y)}^{{\pi}-arcsin(y)} dx dy$
or would I need to make 3 separate integrals?

You need three integrals but you can just do the top half with two integrals and double the answer using symmetry. But remember that using a horizontal "$dy$" element your integral must have the form$$
\int_{y_{smallest}}^{y_{largest}}~x_{right}- x_{left}~dy$$And you can read the $y$ limits from your graph.

I notice that you are giving your answers as double integrals, which has the same effect once you work out the inner integral.

 

[_N3WTON_]

I notice that you are giving your answers as double integrals, which has the same effect once you work out the inner integral.

yes, a double integral is just a definite integral with an integral in the integrand, no?

 

[_N3WTON_]

You need three integrals but you can just do the top half with two integrals and double the answer using symmetry. But remember that using a horizontal "$dy$" element your integral must have the form$$
\int_{y_{smallest}}^{y_{largest}}~x_{right}- x_{left}~dy$$And you can read the $y$ limits from your graph.

Thank you, I've done enough calc for one night but ill give this a shot tomorrow and post my solution