Integrate regions: y = sinx, y = x, x = pi/2, x = pi

[MarkSA]

Hello,

We're currently doing integration on areas between two or more functions, and i'm having trouble with it.

These are the instructions for most of the questions in the homework:
"Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Find the area of the region."

I was wondering if you could explain the general procedure behind this? Here are two of the problems I attempted, but which I don't think are correct.

1) y = sinx, y = x, x = pi/2, x = pi
I did: integral from pi/2 to pi of: (x - sinx)dx
= (1/2)*x^2 + cosx | pi/2 to pi
= [pi^2/2 - 1] - [pi^2/8 + 0] = -pi^2/2 - 1 = -5.9348.
Negative? I didn't think that could be right. I'm also kind of confused about which of those functions is considered to be the 'top' function, since the area between the function is the integral of f(x)-g(x). On the graph it looks like they touch each other.

2) x = y^2 - 4y, x = 2y - y^2. from y = 0 to y = 3
I did: integral from 0 to 3 of (y^2 - 4y - 2y + y^2)dy
= integral from 0 to 3 of (2y^2 - 6y)dy
= (2/3)*y^3 - 3y^2 | 0 to 3
= [18 - 27] - [0 - 0] = -9? Another negative

 

[galactus]

I think you made a mistake on the first one.

\(\displaystyle \L\\\int_{\frac{\pi}{2}}^{\pi}[x-sin(x)]dx=\frac{3{\pi}^{2}}{8}-1\approx{2.701101....}\)

 

[MarkSA]

Ahh you're right. I messed up the very end of that one. The calculator gets the best of me again. Seems like half of my troubles come from entering things into the TI-89 wrong.

Do you know what might be the issue in the second problem?

BTW: Do you know if there is another easy way to tell which function is the 'top' function without graphing it?

 

[galactus]

BTW:

Do you know if there is another easy way to tell which function is the 'top' function without graphing it?

Enter in your values, which gives the larger answer?.

As for the last problem, Also, try solving for y and integrating wrt x. See if you get the same thing. try reversing your functions. You should get 9.

 

[MarkSA]

Thanks. Making x=2y-y^2 the top function gives me the correct answer 9.

I had one other question about this problem... you mentioned solving for y. This is probably bad considering i'm in calculus, but how would you do that? I actually tried it earlier, since I could graph a y= equation easily on my calculator, but couldn't think of a way. For the top function:
x = 2y - y^2
I could do..
x = y(2 - y)
y = x/(2-y)

Is that what you meant? But I have y's on both sides of the equation still. TI-89 can't graph that, at least not in the mode i'm using it in. Unfortunately i'm a little fuzzy on how i'd graph the x= equation by translations (aside from plotting points) too.

 

[galactus]

Forget about solving for y. It's too much trouble and you have the right answer.

The thing to do to solve \(\displaystyle \L\\x=2y-y^{2}\) for x is rewrite it as:

\(\displaystyle \L\\1-(y-1)^{2}\)

\(\displaystyle \L\\x=1-(y-1)^{2}\)

\(\displaystyle \L\\1-x=(y-1)^{2}\)

\(\displaystyle \L\\\sqrt{1-x}=y-1\)

\(\displaystyle \L\\\fbox{\sqrt{1-x}+1=y, \;\ 1-\sqrt{1-x}=y}\)

Here it is anyway:

\(\displaystyle \L\\\int_{0}^{-3}\left[\sqrt{x+4}-2-(\sqrt{1-x}+1)\right]dx=9\)