Double integral: Two lines and one curve

[MSG100]

My problem:

I don't know how to begin with this problem.

 

[Mark44]

My problem:

View attachment 64728

I don't know how to begin with this problem.

Are you sure you wrote the problem correctly? What's the exact problem statement?

The curve y = x² and the line y = x + 2 define a region, but I don't see that y = x², y = x + 2, and x = 4 define a region.

BTW, your graph isn't very useful without the x- and y- axes.

 

[MSG100]

The question should be correct stated. It's translated from another language but I did as good as I could.

Here's a picture of the area.

Should the boundaries maybe be 2 to 4 and (x+2) to x^2?
I haven't done any similar problem with fractions.

 

[Mark44]

OK, I get it now. I had my vertical line in the wrong place. The region as described does make sense. It is the almost triangular shaped region above the line y = x + 2, to the right of the parabola, and to the left of the vertical line x = 4. It would be simpler to integrate with respect to y first, where x + 2 ≤ y ≤ x², and then with respect to x, where 2 ≤ x ≤ 4.

 

[MSG100]

It's late where I am (Europe) so I start to feel sleepy but I have decide to solve this problem first.

I don't seem to get the right answer.

 

[MSG100]

Can anyone find out what I done wrong?

 

[SteamKing]

If I had to guess, I would say that you have the order of integration reversed, at least based on the limits you are showing. For example, in the inner integral, the limits go from x+2 to x^2, both of which are equal to y, yet you are integrating with respect to x. Likewise, for the outer integral, you are integrating w.r.t. y and the limits are from 2 to 4. Check your sketch in Post #3. Clearly in this sketch, the upper y limit is much greater than 4.

 

[MSG100]

I'm confused. If you can write how to begin maybe I can see what I doing wrong. I don't get it.
We just started with double integrals so I need a lot of practice.

 

[Mark44]

Your first integral has the dx and dy in the wrong order. It should be
$$\int_{x = 2}^4 \int_{y = x + 2}^{x^2}\frac{x + 1}{(y - 1)^2} dy~dx$$

It looks like you actually did the integration the right way. In your second integral, you lost dy but haven't actually done the integration yet, so dy should still be present. All you did was put parentheses around the inner integral.

The work in the 3rd integral (2nd line) looks OK, but you need parentheses around that factor of x + 1. Parentheses are also needed around x + 1 in the 4th integral, but otherwise things look correct.

Your mistake is in the 5th integral (3rd line). Check your algebra again on the rational expression in the 4th integral.

 

[MSG100]

I get the same result as before.

Even checked it on wolfram:
http://www.wolframalpha.com/input/?i=%28x%2B1%29*%28%281%2F%281-x^2%29-%281%2F%281-%28x%2B2%29%29

 

[Mark44]

I'm getting 2 - ln(3) as well. Does your book have a different answer? If so, make sure that the problem you showed here is the same as the one in the book. If they are the same, it's possible that the answer in the book is wrong. That has been known to happen.

Part of our confusion here, at least for me, is that some of the things you wrote are different from what you actually did. I mentioned earlier that you had switched dx and dy in the first integral. You also have dy in the 5th integral when you should have had dx. It's also more difficult for us to catch things when we have to open up a separate window to see the work. Having the work all in one place makes it easier for us to find difficulties.

 

[MSG100]

I understand. It's messy and hard to follow my calculations. Hopefully I will learn to write in latex soon.

I don't have the answer but it seems too small, just 0.9 a.u.?

Thanks a lot for the effort you put into helping people like me. Maybe someday we'll be the ones that help others!

 

[Mark44]

I understand. It's messy and hard to follow my calculations. Hopefully I will learn to write in latex soon.

Here's a link to some tips, especially #2 on the list. LaTeX isn't very hard to learn, and you can always use the tags that are available when you click Go Advanced just below the text input area.
https://www.physicsforums.com/showthread.php?t=617567

I don't have the answer but it seems too small, just 0.9 a.u.?

Thanks a lot for the effort you put into helping people like me. Maybe someday we'll be the ones that help others!

You're very welcome. That's our hope as well, that newcomers will eventually become members who can help the ones coming later.