Thread: 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.

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

$\displaystyle \frac{\pi}{18}\int_{1}^{145}{\sqrt{w}}dw=\frac{(14 5\sqrt{145}-1){\pi}}{27}$

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?

I used u-substitution

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)
$\displaystyle \frac{\pi}{27} (1+9x^4)^{3/2} \big|^2_0$

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

$\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)
$\displaystyle \frac{\pi}{27}(1+9(2)^4)^{3/2}=\frac{\pi}{27}(1+9\cdot 16)^{3/2}=\frac{\pi}{27}145^{3/2}$
Okay, do you remember radicals?
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.

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?

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,
$\displaystyle \frac{\pi(145\sqrt{145}-1)}{27}$

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

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.