Finding Area by an Iterated Integral

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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?
 
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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...
 
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.
 
BvU said:
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}
 
BvU said:
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?
 
_N3WTON_ said:
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.
 
LCKurtz said:
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?
 
LCKurtz said:
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
 
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