Need help for text tomorrow

• Aug 1st 2006, 02:59 PM
c_323_h
Need help for text tomorrow
Find the surface area of the solid of revolution over the interval

1.$\displaystyle y=x^3, [0,2]$
Answer: $\displaystyle \pi(145\sqrt{145}-1)/27$

What I tried:

$\displaystyle \frac{dy}{dx}=3x^2$
Plug into formula for Surface Area of a solid of revolution about the x-axis:
$\displaystyle \int_a^b 2\pi y \sqrt{1+(\frac{dy}{dx})^2}dx$
$\displaystyle 2\pi\int_0^2 x^3\sqrt{1+(3x^2)^2}dx$
$\displaystyle 2\pi\int_0^2\sqrt{1+9x^4}x^3dx$
let $\displaystyle u=1+9x^4, \frac{du}{dx}=36x^3$
Convert limits: when$\displaystyle x=0, u=1$, when $\displaystyle x=2, u=289$
$\displaystyle \frac{2\pi}{36}\int_1^{289} \sqrt{u} du$
Integrate:
$\displaystyle \frac{\pi}{18}(\frac{2}{3}u^{\frac{3}{2}})\bigg|^{ 289}_1$
Plug in $\displaystyle u$, evaluate at limits, simplify.

I don't seem to be getting the answer that's in my book.
• Aug 1st 2006, 03:37 PM
galactus
Let $\displaystyle u=x^{4};\;\ \frac{du}{4}=x^{3}dx$

Then you get:

$\displaystyle \frac{\pi}{2}\int_{0}^{16}\sqrt{1+9u}du$

Let $\displaystyle w=1+9u;\;\ \frac{dw}{9}=du$

$\displaystyle \frac{\pi}{18}\int_{1}^{145}{\sqrt{w}}dw=\frac{(14 5\sqrt{145}-1){\pi}}{27}$
• Aug 1st 2006, 03:41 PM
c_323_h
Quote:

Originally Posted by galactus
Let $\displaystyle u=x^{4};\;\ \frac{du}{4}=x^{3}dx$

Then you get:

$\displaystyle \frac{\pi}{2}\int_{0}^{16}\sqrt{1+9u}du$

thanks. how do you integrate this without using substitution? shouldn't either way work though?
• Aug 1st 2006, 03:48 PM
galactus
I used u-substitution
• Aug 1st 2006, 06:05 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Find the surface area of the solid of revolution over the interval

1.$\displaystyle y=x^3, [0,2]$
Answer: $\displaystyle \pi(145\sqrt{145}-10\sqrt{10})/27$

I shall use the infamous earboth substitution.

$\displaystyle 2\pi \int_0^2 x^3\sqrt{1+[(x^3)']^2}dx$
Thus,
$\displaystyle 2\pi \int_0^2 x^3\sqrt{1+(3x^2)^2} dx$
Thus,
$\displaystyle 2\pi \int_0^2 x^3\sqrt{1+9x^4} dx$
Express as, (and you will see why),
$\displaystyle \frac{2\pi}{36} \int 36x^3 \sqrt{1+9x^4} dx$
Let $\displaystyle u=1+9x^4$
Then, $\displaystyle \frac{du}{dx}=36x^3$
Thus,
$\displaystyle \frac{\pi}{18} \int \sqrt{u} \frac{du}{dx} dx$
Which is (after earboth substitute),
$\displaystyle \frac{\pi}{18} \int \sqrt{u} du$
Thus,
$\displaystyle \frac{\pi}{18} \cdot \frac{2}{3} u^{3/2}$
Thus, (substitute back)
$\displaystyle \frac{\pi}{27} (1+9x^4)^{3/2} \big|^2_0$
• Aug 1st 2006, 06:28 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
Thus, (substitute back)
$\displaystyle \frac{\pi}{27} (1+9x^4)^{3/2} \big|^2_0$

How come you didn't have to convert the limits? do you mind evaluating this for me? I know it's really simple...I can get this far but when I evaluate I don't get the same answer as in the book:
$\displaystyle \pi(145\sqrt{145}-1)/27$

I know that they factored...but don't quite exactly see how they got there. When I evaluate I get huge numbers. Thanks
• Aug 1st 2006, 06:44 PM
ThePerfectHacker
$\displaystyle \frac{\pi}{27} (1+9x^4)^{3/2} \big|^2_0$

This is how I taught my students, since they get confused with negatives and all that.

2 seperate things.
1.1)1]1}First: Evaluate it for upper and stop.
2.2)2]2}Second: Evaluate it for lower and stop.

That is it now subtract them. The reason why I prefer it like this is because it elimanates much effort with working it all out in one long line which confuses many students.
---
Step 1)
$\displaystyle \frac{\pi}{27}(1+9(2)^4)^{3/2}=\frac{\pi}{27}(1+9\cdot 16)^{3/2}=\frac{\pi}{27}145^{3/2}$
That nasty exponent means,
$\displaystyle \sqrt{145^3}=\sqrt{145^2\cdot 145}=145\sqrt{145}$
Thus, the first part gives,
$\displaystyle \frac{\pi\cdot 145}{18} \cdot \sqrt{145}$

Okay, this is the problem are you sure that the limits are 0 and 2. They should be 1 and 2 to give the book's answer.

You see how to do it? Do you want to do the second part. I just really do not want to do it because of all the time it takes.
• Aug 1st 2006, 06:54 PM
c_323_h
sigh..i posted the wrong answer. this is the second time this week. :( it's supposed to be
$\displaystyle \pi(145\sqrt{145}-1)/27$

it is over the correct interval though. when you used u-substition why didn't you have to convert the limits?
• Aug 1st 2006, 07:00 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
sigh..i posted the wrong answer. this is the second time this week. :( it's supposed to be
$\displaystyle \pi(145\sqrt{145}-1)/27$

it is over the correct interval though

Now it makes sense. And let me continue.

Now you do step 2 you evaluate it for 0 only:
$\displaystyle \frac{\pi}{27}(1+9(0)^4)^{3/2}=\frac{\pi}{27}(1+9\cdot 0)^{3/2}=\frac{\pi}{27}1^{3/2}$
Okay, do you know that,
$\displaystyle 1^n=1$ for any $\displaystyle n$?
Then, you are left with,
$\displaystyle \frac{\pi}{27}$.
---
In my other post I did the first equaluation
In this post I did the second equaluation.
Now subtract,
$\displaystyle \frac{\pi\cdot 145}{18} \cdot \sqrt{145}-\frac{\pi}{27}$
Fractions, common denominator good,
$\displaystyle \frac{\pi \cdot 145\sqrt{145}-\pi}{27}$
Factor pi,
$\displaystyle \frac{\pi(145\sqrt{145}-1)}{27}$
• Aug 1st 2006, 07:20 PM
c_323_h
ok. i understand that you used 2 and 0 as the limits.

general question on definite intergrals: when you use u-substitution, do you always convert the limits, or can you treat the definite integral as an indefinite integral and evaluate the original limits at the end? which would be 0 and 2 in this case
• Aug 1st 2006, 07:27 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
ok. i understand that you used 2 and 0 as the limits.

general question on definite intergrals: when you use u-substitution, do you always convert the limits, or can you treat the definite integral as an indefinite integral and evaluate the original limits at the end? which would be 0 and 2 in this case

The method I prefer is doing it like an indefinite integral and the converting back. So you use the original limits.

You can also, change the limits with a "u-substitution" but I do not like that one.