# Calc II

1. Jan 15, 2007

### cscott

1. The problem statement, all variables and given/known data

Find the volume of the solid which is obtained by revolving the discs $$(x - 1)^2 + y^2 \le 1$$

3. The attempt at a solution

I say using cylinders would be easiest so,
$$\int_a^b{2\pi r(x)h(x) dx} = \int_0^2{2\pi x\sqrt{1-(x-1)^2} dx}$$

Any hints on how to get started with this integral?

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1. The problem statement, all variables and given/known data

Find the arclength of the curve $y = e^x$ with $0 \le x \le ln 2$

3. The attempt at a solution
$$y' =e^x$$
so I need to solve
$$\int_a^b{\sqrt{1+(y')^2}dx} = \int_0^{ln 2}{\sqrt{1+e^{2x}}dx}$$

Any hints?

Last edited: Jan 16, 2007
2. Jan 15, 2007

### sara_87

$$\int{\sqrt{1+e^{2x}}$$

this can be written as $${1+e^{2x}}^(1/2)$$

then to integrate you add one to the power...etc

does this help?

3. Jan 15, 2007

### sara_87

wait that came up wrong i meant:

this can be written as (1+e^(2x))^(1/2)

4. Jan 16, 2007

### cscott

I can't do what you said as it's a composite function...

Any other ideas?

5. Jan 17, 2007

### Gib Z

sara_87, when doing exponents longer than one character, uses { } instead of ( ). It seems you just forgot, since you did it the first time with the 2x :D

6. Jan 17, 2007

### Gib Z

o, cscott, would It help If You had it in this form? (Hopefully you know integration by parts ).

Make a u substitution
$$u=e^{2x}$$
$$\frac{du}{dx}=2e^{2x}$$
Solve for dx
$$dx=\frac {du}{2e^{2x}}=\frac{du}{2u}$$
Now you just need to do the rest.

$$\int (u+1)^{\frac{1}{2}} 2u du$$

Have fun :D

Last edited: Jan 17, 2007
7. Jan 17, 2007

### Schrodinger's Dog

question 1)Not that I want to ruin your fun here, but by revolving discs do you mean a sphere? If I draw a graph of a revolving disc unless it's simply rotating about it's axis which gives a circle, wouldn't I end up with a sphere? in other words $$\frac{4}{3}\pi r^3$$

Last edited: Jan 17, 2007
8. Jan 17, 2007

### cristo

Staff Emeritus
You made a slight mistake; the integral is $$\int \frac{(u+1)^{\frac{1}{2}}}{2u} du$$

9. Jan 17, 2007

### cscott

Sorry, it should say "revolve the disc defined by ___ around the y-axis" which I think calls for cylinders to make it easiest and wouldn't be a sphere.

Last edited: Jan 17, 2007
10. Jan 17, 2007

### Schrodinger's Dog

Oh you mean revovle a cyclinder around the y axis. I suggest you copy out the question again exactly as it is on paper, as your original question is a little misleading. I thought you meant spinning a circle through 360 degrees.

11. Jan 17, 2007

### Gib Z

cristo- ty for spotting this, my brain seems to be farting alot recently...

cscott- either way, I still think integration by parts helps

12. Jan 18, 2007

### Schrodinger's Dog

The first question I'm assuming your talking about a toroidal. which is found from the curve y=e^x. You may want to look up toroidal volumes.

Hint: you should end up with the second question integral, part 1 if you have done it correctly.

i.e. $$\int_a^b{\sqrt{1+(y')^2}dx= \int_0^{ln 2}{\sqrt{1+e^{2x}}dx}$$

The second I can only say was too complicated for the level I'm at although once I saw how it was done I could follow the steps easilly enough(I asked Dr.E.P.A Bailey. A colleague to solve it basically) Otherwise known as cheating

let $$u^2=1+e^{2x}$$ and then substitute.

then put in the form of partial fractions.

$$\frac {f(u)}{g(u)}$$

where f(u) and g(u) are polynomials.

You should end up with this:-

$$\int_0^{ln 2} 1+\frac {1}{2(u-1)}. \frac {-1}{2(u+1)}$$

from here it should be fairly straightforward if you express the answer in the form of natural logs.

Hope that helps.

Last edited: Jan 18, 2007