# Area and Volume problems

1. Jul 14, 2007

### giant016

I'm having some trouble on all of these problems.

For the first one, I don't know how to do it because of the two variables (although they're both the same).

For #2, I figure to use the disk method and set the integral from 5 to 13. The only problem is when I substitute 5 and 13 in for x and subtract them after I've integrated, I come up with a negative number.

For #3 and 4 the problem I have is that I'm trying to integrate on the interval of x=0 to x=9. Where x=9 however, the formula comes out to 0, so the volume does as well. Once I get #3, #4 should be cake.

Any guidance is much appreciated.

2. Jul 14, 2007

### olgranpappy

The first problem is pretty simple. You should draw a picture. There's no other way to understand the problem. Draw a set of x and y axes. And then figure out what those three different regions and lines mean. For example, the fact that 0<x<1 means I only care about an infinite strip which runs parallel to the y-axis and has a thickness of 1 with one edge at x=0 and one edge at x=1. Shade that region on your picture.

Now, the equation "y=0" is the equation for a line. What does that line look like? where do you draw it on your picture?

Similarly the equation "y=xe^-x" is a line. What does it look like. Draw it on your picture.

3. Jul 15, 2007

### rocomath

here's problem #2 (disk and shell method), but if my answer is wrong, then i'm a failure :)

http://img214.imageshack.us/img214/4121/volumero5.jpg [Broken]

idk how to do the others cuz i've only taken Calc 1, so hope someone else will be nice enough to help you. gl!

(just looking at the problem, looks like you have it in y = mx + b form)

lol ok out of curiousity, i tried it but ... idk if i even took the right approach ...

graph = calculator = http://img149.imageshack.us/img149/8840/volumelc9.jpg [Broken]

take it ez if i'm wrong, i only have 6 weeks of Calculus under my belt, lol.

Last edited by a moderator: May 3, 2017
4. Jul 15, 2007

### HallsofIvy

Staff Emeritus
If you are expected to be able to do 3 or 4, 1 should be trivial. From the basic definition of the integral, the area under the graph of y= f(x), above y= 0 and from x= a to x= b is
$$\int_a^b f(x) dx$$.
(olegranpappy might be just a little misleading. When he says "the equation is a line" he means that the graph of the function is a curve, not a straight line.)

As for 3 and 4, draw the graph and use "symmetry"- if you can find the volume of part of the graph, symmetry gives you the volume of the whole thing.

5. Jul 15, 2007

### olgranpappy

Yeah, a line--not a straight line--just a line.

the equation y=0 gives you a straight line, the equation y=xe^-x gives you a not-straight line...

6. Jul 15, 2007

### giant016

The thing that is throwing me off is that x is in both variables. Do I just integrate them seperately and multiply them. i.e. (x^2)/2 times e^-x? I know I can't bring x or e^-x out to the front of the integrand like a constant.

Also, how are you guys typing interals and stuff out?

7. Jul 15, 2007

### Schrodinger's Dog

The latex format of maths.

$$\int_a^b e^{x^x}dx$$

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = \lim_{n\rightarrow\infty} (1+x/n)^n$$

You can click on any latex script and it'll show you how it's done.

Last edited: Jul 15, 2007
8. Jul 15, 2007

### olgranpappy

No. That would not work. Why should it? The only way to do the integral is to do it correctly.

Have you been going to class or studying your book at all? That would probably help a lot.

So, let's see here... do you know how to evaluate this integral:
$$\int e^{-x}dx$$
and do you know the technique of "integration by parts?"

If you do know these things then you will be able to solve part (1). Else, there is little hope. If you don't know these things, we can help you learn them, but you will have to work at it. Cheers.

9. Jul 15, 2007

### giant016

Ah...how did I not think of that? I did try it, but came up with a negative answer. It seems like it should work out, so could somebody point out where I messed up?

Here's where I'm stuck on #3 as well. When I substitute 9 back in for x I get 0/4. Any hints?

10. Jul 15, 2007

### olgranpappy

...the derivative of $$e^{-x}$$ is not $$e^{-x}$$, it's
$$-e^{-x}$$.

11. Jul 15, 2007

### olgranpappy

p.s. for part one your answer must be greater than zero, and it happens to be less than one.

12. Jul 16, 2007

### HallsofIvy

Staff Emeritus
You really need to go back and review basic integration and differentiation! Your idea that you can integrate a product by integrating the factors separately tells us that!

As olgranpappy said, the derivative of e-x is -e-x- don't forget the "chain rule".

In problem 3,
$$\int (9^\frac{2}{3}- x^\frac{2}{3})^3 dx[/itex] is not [tex]\frac{(9^\frac{2}{3}- x^\frac{2}{3})^4}{4}$$
You forgotten the power of x inside.

Since there is no (2/3)x-1/3 to use for a "substitution", you will need to actually multiply (92/3- x2/3)3 out.

Of course, once you have done 3, 4 is trivial- think "symmetry".

13. Jul 16, 2007

### Schrodinger's Dog

Yeah you've got the wrong integral for g'(x)

Integration by parts:-

$$\int {f(x)g'(x)}\ dx\ =\ f(x)g(x)\ -\int {f'(x)g(x)\ dx$$

f(x)=x
g'(x)=-e-x
f'(x)=1
g(x)=e-x

$$\rightarrow -xe^{-x}\ -e^{-x}(1)\ = -e^{-x}(x+1)+c$$

$$\int_0^1xe^{-x}\rightarrow(-0.368-0.368)-(0-1)=0.264$$

Last edited: Jul 16, 2007