Changing order of integration

1. Apr 22, 2016

says

1. The problem statement, all variables and given/known data
Switch the order of integration:
∫∫f(x,y) dydx, with 0≤x≤π/2, 0≤y≤sin(x)

3. The attempt at a solution
∫∫f(x,y) dxdy, with sin(x)≤x≤π/2, 0≤y≤sin(x)

The upper bound of x is π/2, while the lower bound is sin(x). It's just the first 1/4 of the area of a sine wave. The upper bound of y is sin(x) and lower bound is 0.

I was just wondering if someone could confirm. I've been having a bit of trouble with integrals and assigning regions. Thank you :)

2. Apr 22, 2016

Staff: Mentor

No. In the first integral above, you are looking at vertical strips that go from y = 0 to y = sin(x), and summing them as x ranges from x = 0 to x = pi/2.
No. Switching the order of integration means that you will now be using horizontal strips, and summing them from bottom to top.
What is the leftmost x-value on a given horizontal strip? (The rightmost x-value is pi/2.)
You need to be thinking about inverse functions.

3. Apr 22, 2016

says

The rightmost value is pi/2, so the leftmost value would be sin(x), no?

4. Apr 22, 2016

says

Or should I be saying the leftmost value is 0?

I'm saying sin(x) because when we sub in sin(0) we get 0.

5. Apr 22, 2016

LCKurtz

Think of a little dydx square in the middle of your area. Now move it left and right. Your inner integral should look like $\iint_{x_{left}}^{x_{right}}f(x,y)~dxdy$. When you move that dydx square to the left it hits the sine curve. What is the x value there in terms of y? The value sin(x) you suggested in post #3 is the y value, not the x value.

6. Apr 22, 2016

Staff: Mentor

Have you drawn a sketch of the region over which integration is being performed? If not, you really should.

Also, instead of saying "leftmost value," it's better to specify leftmost x value, as well as lower y value and upper y value.

7. Apr 22, 2016

says

I drew a picture. Always the first thing I do.

The leftmost x value, judging by my picture, = 0, while the rightmost value = pi/2
The upper y value = sin(x) while the lower y value = 0

0≤x≤pi/2
0≤y≤sin(x)

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8. Apr 22, 2016

Staff: Mentor

For the shaded region, is the leftmost x value 0 for each point on the left edge of the region?
Look at what LCKurtz wrote a few posts back.

9. Apr 22, 2016

says

opps, please refer to post #7 :)

10. Apr 22, 2016

LCKurtz

Which makes me think you haven't read my post #5.

11. Apr 22, 2016

says

Oh, so I want the leftmost value for the whole region. I was just writing down 0 because 0 is the leftmost point!

12. Apr 22, 2016

says

The leftmost x=sin(x) and the rightmost pi/2
the upper y = sin(x) and lower = 0

How is this different from my original answer though?

sin(x)≤x≤pi/2
0≤y≤sin(x)

13. Apr 22, 2016

Staff: Mentor

The two inequalities should really be 0 ≤ y ≤ sin(x), 0 ≤ x ≤ π/2
In the original integral, you are using vertical strips, each of which ranges from y = 0 to y = sin(x). The outer integral sums all of these vertical strip areas, as x ranges between 0 and π/2.

Switching the order of integration means you need a different description of the region. In this integral, the inner integral uses horizontal strips, each of which ranges from x = $x_{left}$ to x = $x_{right}$. The horizontal strip areas are then summed as y ranges from 0 to 1.

You need to figure out what $x_{left}$ and $x_{right}$ are. I'm pretty much repeating what I said in post #2.

Note: I've been a little sloppy by saying "area" above. Since the integrand contains a function, you aren't actually computing the area.

14. Apr 22, 2016

Staff: Mentor

x = sin(x) makes no sense here.

For the equation y = sin(x), if you know an x value, it's easy to get the y value. However, what if you know the y value? How do you get the x value on the same graph?
It's not different -- that's the problem, since your first answer was wrong.

15. Apr 22, 2016

says

0 ≤ y ≤ sin(x), 0 ≤ x ≤ π/2

if x=0 then sin(0) = 0, so how come sin(x)≤x≤pi/2 can't be a valid region also?
I don't understand what you mean by saying x = sin(x) makes no sense?

16. Apr 22, 2016

says

Ignore that last question! I understand now. Thanks :)

17. Apr 22, 2016

LCKurtz

18. Apr 22, 2016

says

∫∫f(x,y) dxdy, with 0 ≤ x ≤ π/2, 0 ≤ y ≤ sin(x) being the lower and upper bounds of integration.

19. Apr 22, 2016

Staff: Mentor

No. All you really have done is switch dx and dy in the integral, and swap the order of the inequalities. Go back and read the replies in posts 2, 5, 10, 12, 15, and 16.

20. Apr 22, 2016

says

It's just that one lower bound x value I'm confused with.

arcsin?