Math Help - Area and rotation volume

1. Area and rotation volume

a) Calculate the are between the lines $y=x$ and $y^2=2-x$ and the parabola $y^2=2-x$

b) Calculate the rotation volume of the body that comes when the curve $y=xe^x$ rotates around the x-axis

Is there a way to post a coordinate system here with curves and stuff in it?

2. Originally Posted by lynch-mob
a) Calculate the are between the lines $y=x$ and $y^2=2-x$ and the parabola $y^2=2-x$
First find the two intersections of the curves (simple algebra). Then integrate the difference between the two curves (i.e. $\int_a^b\left\lvert f(x) - g(x)\right\rvert\,dx$). You will have to be careful here that you are subtracting the right functions on each interval: I recommend sketching a graph. It will help if you split the second equation into $y = \pm\sqrt{2-x}$.

Originally Posted by lynch-mob
b) Calculate the rotation volume of the body that comes when the curve $y=xe^x$ rotates around the x-axis
Use the disc method $\left(V = \pi\int_a^bf(x)^2\,dx\right)$ to find the volume. In this case you can integrate by parts, but you will need to do it twice.

Originally Posted by lynch-mob
Is there a way to post a coordinate system here with curves and stuff in it?
Yes. Create the graph in an external program and save it in an image format. Then upload it using the "Manage Attachments" button when making your post.

3. Mmhm

Originally Posted by Reckoner
First find the two intersections of the curves (simple algebra). Then integrate the difference between the two curves (i.e. $\int_a^b\left\lvert f(x) - g(x)\right\rvert\,dx$). You will have to be careful here that you are subtracting the right functions on each interval: I recommend sketching a graph. It will help if you split the second equation into $y = \pm\sqrt{2-x}$.
By the way the second function was $y=2-x$ and it´s actually the negative side of the parabola that I should use. Sorry for the mess up!

Ok so i find the intersection points by putting $x^2=2-x$ and $(2-x)^2=2-x$
When I get the x-coordinates I can easily find the y-coordinates. So I have the intersections (1,1),(2,0) and (-2,-2). So basically the x-values will be the intervals I´ll be using?

Somebody told me to divide the figure in two by drawing a vertical line crossing (1,1) in order to more easily calculate the two areas.Is this right?

$\int_1^2 ((2-x)-(-\sqrt(2-x))dx$

This equals $\bigg[2x-\frac{1}{2} x^2-\frac{2}{3}(2-x)^\frac{3}{2}\bigg]_1^2$

And then i put in the interval values, first 2 and subtract the result with the result when putting in 1?

which is $\frac{7}{6}$ ??

The second area is calculated in similar fashion?

Which would be $\frac{19}{6}$

Added together $\frac{13}{3}$?

Originally Posted by Reckoner
Use the disc method $\left(V = \pi\int_a^bf(x)^2\,dx\right)$ to find the volume. In this case you can integrate by parts, but you will need to do it twice.
I have $V = \pi\int_a^b(xe^x)^2\,dx$ but which values should I put for a and b? And how to integrate that? I´m so stuck with this. I use the formula $\int f(g(x))=F(x)g(x)-\int F(x)g\prime(x)$ but get nowhere.

Originally Posted by Reckoner
Yes. Create the graph in an external program and save it in an image format. Then upload it using the "Manage Attachments" button when making your post.
What kind of free software would you recommend?

4. Originally Posted by lynch-mob
By the way the second function was $y=2-x$ and it´s actually the negative side of the parabola that I should use. Sorry for the mess up!
Okay, let me make sure I understand what problem you are dealing with. You want the area of the region bounded by the three curves $y = x$, $y = 2 - x$, and $y^2 = 2 - x$. Is that correct? In this case, you would be correct that you only need the negative side of the parabola, as it can be shown that, on the intervals of interest, those two lines lie below the upper half.

Originally Posted by lynch-mob
Ok so i find the intersection points by putting $x^2=2-x$ and $(2-x)^2=2-x$
When I get the x-coordinates I can easily find the y-coordinates. So I have the intersections (1,1),(2,0) and (-2,-2). So basically the x-values will be the intervals I´ll be using?
Those are the correct intersections.

Originally Posted by lynch-mob
Somebody told me to divide the figure in two by drawing a vertical line crossing (1,1) in order to more easily calculate the two areas.Is this right?
It may be helpful to do so. Basically you will have two intervals: for $x\in[-2,\;1],\;-\sqrt{2 - x} \leq x \leq 2 - x$ and for $x\in[1,\;2],\;-\sqrt{2 - x} \leq 2 - x \leq x$. These inequalities will determine which order you subtract the functions when integrating.

Originally Posted by lynch-mob

$\int_1^2 ((2-x)-(-\sqrt(2-x))dx$

This equals $\bigg[2x-\frac{1}{2} x^2-\frac{2}{3}(2-x)^\frac{3}{2}\bigg]_1^2$

And then i put in the interval values, first 2 and subtract the result with the result when putting in 1?

which is $\frac{7}{6}$ ??

The second area is calculated in similar fashion?

Which would be $\frac{19}{6}$

Added together $\frac{13}{3}$?
Bravo! Nice work.

Originally Posted by lynch-mob
I have $V = \pi\int_a^b(xe^x)^2\,dx$ but which values should I put for a and b? And how to integrate that? I´m so stuck with this. I use the formula $\int f(g(x))=F(x)g(x)-\int F(x)g\prime(x)$ but get nowhere.
Don't worry! Integration by parts can be tricky at first, but once you get used to it you can get quite good at it. If you aren't sure at first which values to use for each "part," just take a good guess and see if you wind up with something simple enough to evaluate. If not, try again! You will get the hang of it.

In this case we want $\int\left(xe^x\right)^2\,dx = \int x^2e^{2x}\,dx$.

The formula for integration by parts is $\int u\,dv = uv - \int v\,du$. Here's my suggestion: for $dv$ choose the most complicated portion of the integrand that you know how to integrate. In our case, $e^{2x}$ is the most complicated factor, but it is integrated easily. So we would have

$\begin{array}{rclcrcl}
u & = & x^2 & \Rightarrow & du & = & 2x\,dx\\
dv & = & e^{2x}\,dx & \Rightarrow & v & = & \frac12e^{2x}
\end{array}$

Once you make your substitutions, you will end up with

$uv - \int v\,du = \frac12x^2e^{2x} - \int xe^{2x}\,dx$

This integral, $\int xe^{2x}\,dx$, will require another application of integration by parts.

In general, here's some tips for choosing your $u$ and $dv$:

For integrals of the form $\int x^n\sin x\,dx$ (and other trig functions), let $u = x^n$ and $dv = \sin x\,dx$.

Similarly, for integrals of the form $\int x^ne^x\,dx$, let $u = x^n$ and $dv = e^x\,dx$.

For the integrals $\int e^x\sin x\,dx$ or $\int e^x\cos x\,dx$, let $u = \sin x$ or $u = \cos x$, and let $dv = e^x\,dx$. Use integration by parts twice, and you will actually end up with an equation that can be solved for the desired integral.

For integrals like $\int\ln x\,dx$ or $\int\arcsin x\,dx$, let $u$ be the entire integrand with $dv = dx$.

Originally Posted by lynch-mob
What kind of free software would you recommend?
There's all sorts of stuff out there. If you have a TI calculator and the appropriate cable (I know the 89s have one included) you can use their software to upload screen output directly to your computer.

You can also do some searching around on Google. I found, for example: MathGV and gnuplot.

If you have Windows XP, you can also look at the Windows XP PowerToys. They have a very nice calculator that can do graphing.

And for simple graphs, you can often create them by hand in an image editor.

5. Hi

Originally Posted by Reckoner
Don't worry! Integration by parts can be tricky at first, but once you get used to it you can get quite good at it. If you aren't sure at first which values to use for each "part," just take a good guess and see if you wind up with something simple enough to evaluate. If not, try again! You will get the hang of it.

In this case we want $\int\left(xe^x\right)^2\,dx = \int x^2e^{2x}\,dx$.

The formula for integration by parts is $\int u\,dv = uv - \int v\,du$. Here's my suggestion: for $dv$ choose the most complicated portion of the integrand that you know how to integrate. In our case, $e^{2x}$ is the most complicated factor, but it is integrated easily. So we would have

$\begin{array}{rclcrcl}
u & = & x^2 & \Rightarrow & du & = & 2x\,dx\\
dv & = & e^{2x}\,dx & \Rightarrow & v & = & \frac12e^{2x}
\end{array}$

Once you make your substitutions, you will end up with

$uv - \int v\,du = \frac12x^2e^{2x} - \int xe^{2x}\,dx$

This integral, $\int xe^{2x}\,dx$, will require another application of integration by parts.

In general, here's some tips for choosing your $u$ and $dv$:

For integrals of the form $\int x^n\sin x\,dx$ (and other trig functions), let $u = x^n$ and $dv = \sin x\,dx$.

Similarly, for integrals of the form $\int x^ne^x\,dx$, let $u = x^n$ and $dv = e^x\,dx$.

For the integrals $\int e^x\sin x\,dx$ or $\int e^x\cos x\,dx$, let $u = \sin x$ or $u = \cos x$, and let $dv = e^x\,dx$. Use integration by parts twice, and you will actually end up with an equation that can be solved for the desired integral.

For integrals like $\int\ln x\,dx$ or $\int\arcsin x\,dx$, let $u$ be the entire integrand with $dv = dx$.
I see we use a little diffrent style concerning partial integration. Let me try it my way, I suppose you´ll understand it anyway. The expressions are still the same. I was also wondering if my teacher forgot to put in interval values in the task. Because the graph circling around the x-axis will be a body with an infinite volume. Hmmm??

Anyway here´s my attempt (I´ll leave the $V$ and $\pi$ for later and only try the integration part:

$\int_a^b(xe^x)^2\,dx = \int_a^bx^2e^{2x}\,dx = x^2\frac{1}{2}e^{2x}\,-\,\int\frac{1}{2}e^{2x}2x\,=$

$x^2\frac{1}{2}e^{2x}\,-\,\bigg(\frac{1}{2}xe^{2x}\,-\,\int\frac{1}{2}e^{2x}\cdot1\bigg)$

Which is

$x^2\frac{1}{2}e^{2x}\,-\,\bigg(\frac{1}{2}xe^{2x}\,-\,\frac{1}{4}xe^{2x}\bigg)$

Which is

$x^2\frac{1}{2}e^{2x}\,-\,\frac{1}{4}xe^{2x}$

And then putting it in the formula we get:

$V=\pi \cdot\,\bigg(what\,do\,I\,do\,now?\bigg)$

6. Originally Posted by lynch-mob
I see we use a little diffrent style concerning partial integration. Let me try it my way, I suppose you´ll understand it anyway. The expressions are still the same.
Yes, it's just a different way of expressing the same thing. Use whichever one you are more comfortable with.

Originally Posted by lynch-mob
$x^2\frac{1}{2}e^{2x}\,-\,\bigg(\frac{1}{2}xe^{2x}\,-\,\int\frac{1}{2}e^{2x}\cdot1\bigg)$

Which is

$x^2\frac{1}{2}e^{2x}\,-\,\bigg(\frac{1}{2}xe^{2x}\,-\,\frac{1}{4}{\color{red}x}e^{2x}\bigg)$
Where did that $x$ come from?

You should get

$x^2\frac12e^{2x}-\left(\frac12xe^{2x}-\frac14e^{2x}\right)$

$=\frac12x^2e^{2x}-\frac12xe^{2x}+\frac14e^{2x}$

Originally Posted by lynch-mob
Which is

$x^2\frac{1}{2}e^{2x}\,-\,\frac{1}{4}xe^{2x}$

And then putting it in the formula we get:

$V=\pi \cdot\,\bigg(what\,do\,I\,do\,now?\bigg)$
From before, you have:

$V = \pi\int_a^b x^2e^{2x}\,dx$

$=\pi\left[\frac12x^2e^{2x}-\frac12xe^{2x}+\frac14e^{2x}\right]_a^b$

Now, I do suspect you should have been given some kind of interval. Although you can have a surface of revolution created from an infinitely long curve and still have it enclose a finite volume (see e.g. Gabriel's Horn for a surprising example), this volume would definitely be infinite if you evaluated it over the whole real line.

7. Originally Posted by Reckoner
Yes, it's just a different way of expressing the same thing. Use whichever one you are more comfortable with.
I suppose that has something to do with me coming from Finland. For instance I have never seen the big brackets (after integrating ) before. Over here we use a big slash with the interval values on top and on bottom of the slash. Well, nice to learn new stuff.

Originally Posted by Reckoner
Where did that $x$ come from?
Typo

Originally Posted by Reckoner
From before, you have:

$V = \pi\int_a^b x^2e^{2x}\,dx$

$=\pi\left[\frac12x^2e^{2x}-\frac12xe^{2x}+\frac14e^{2x}\right]_a^b$

Now, I do suspect you should have been given some kind of interval. Although you can have a surface of revolution created from an infinitely long curve and still have it enclose a finite volume (see e.g. Gabriel's Horn for a surprising example), this volume would definitely be infinite if you evaluated it over the whole real line.
Ok. I think I´ll just leave it like that. All i need to do is put in the interval values later on and do some subtraction in case he gives me an interval.

You think you can help me with another thing. Since I write all my homework on a computer it would be nice to write it with some fancy software. I´ve been using word2007 which is okay but there are some features I miss. Like the big brackets for instance. This Latex thing which is totally new to me is superb. Im in love with it. I mean I can do basically anything with it, well not yet but after I´ve used it for a while
Word 2007 would be excellent if I could "insert latex box" where ever in the document.

-Lynch

8. Originally Posted by lynch-mob
I suppose that has something to do with me coming from Finland. For instance I have never seen the big brackets (after integrating ) before. Over here we use a big slash with the interval values on top and on bottom of the slash. Well, nice to learn new stuff.

Typo

Ok. I think I´ll just leave it like that. All i need to do is put in the interval values later on and do some subtraction in case he gives me an interval.

You think you can help me with another thing. Since I write all my homework on a computer it would be nice to write it with some fancy software. I´ve been using word2007 which is okay but there are some features I miss. Like the big brackets for instance. This Latex thing which is totally new to me is superb. Im in love with it. I mean I can do basically anything with it, well not yet but after I´ve used it for a while
Word 2007 would be excellent if I could "insert latex box" where ever in the document.

-Lynch
Vista has a fantastic equation editor if you are serious about math, it mitigates all my math type problems when typing a long math document

9. Originally Posted by lynch-mob
You think you can help me with another thing. Since I write all my homework on a computer it would be nice to write it with some fancy software. I´ve been using word2007 which is okay but there are some features I miss. Like the big brackets for instance. This Latex thing which is totally new to me is superb. Im in love with it. I mean I can do basically anything with it, well not yet but after I´ve used it for a while
Word 2007 would be excellent if I could "insert latex box" where ever in the document.
Well, if you like you can write your entire paper in $\text{\LaTeX}$ (it produces beautiful output even for things besides mathematics--though the difference probably isn't worth it for things like English papers).

Download a $\text{\TeX}$ port: MiKTeX is a good one; you can find others at the CTAN (they also have a nice introduction and help for beginners).

If submitting your papers online, you can use pdfTeX (included with most distributions) to get a PDF file.

It may take you a little while to get up and running, and your writing might be a little slow until you get the hang of it, but it is worth it if you want professional, quality output.