# Thread: Definite intergral

1. ## Definite intergral

Hi,
I am solving a problem which involves calculating the surface are of a solid revolution about the x – axis. I have found that the surface area can be given by:

$\displaystyle A = 2\Pi \int_{-70}^{110}r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx$

I have done this using the trapezium rule as I know the values of r and d, so I have a rough estimate of a figure I am aiming for.
The question I am trying to answer says to use the standard integral:

$\displaystyle \int \frac{dx}{\sqrt{k^{2}-x^{2}}}= \tan^{-1}\left ( \frac{x}{\sqrt{k^{2}-x^{2}}} \right )+C$

I am struggling to interpret the r+rd/√(r^2-x^2) into the form dx/√(k^2-x^2) in order to evaluate the integral. Any help or guidance would be appreciated as I have spent too much time trying to do this.

Thanks

2. ## Re: Definite intergral

$\displaystyle \frac{d}{dx}\left (\arcsin{x} \right) = \frac{1}{\sqrt{1-x^2}}$

Assuming $\displaystyle r > 0$:
$\displaystyle \frac{r}{\sqrt{r^2-x^2}}=\frac{1}{\sqrt{1-(x/r)^2}}$

This is very close to the derivative posted above. What is the derivative of $\displaystyle \arcsin(x/r)$?

3. ## Re: Definite intergral

$\displaystyle \int_{-70}^{110}r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx$

$\displaystyle = \int_{-70}^{110}r \: dx+ \int_{-70}^{110}\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx$

$\displaystyle = r\int_{-70}^{110}1 \: dx + rd \int_{-70}^{110}\frac{1}{\sqrt{r^{2}-x^{2}}}\: dx$

Is this what you mean?

4. ## Re: Definite intergral

Hello, freethinker!

I am solving a problem which involves calculating the surface area of a solid revolution about the x–axis.

I have found that the surface area can be given by: .$\displaystyle A \:=\:2\pi \int_{-70}^{110}\left(r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\right)\,dx$

I have done this using the trapezium rule as I know the values of r and d,
so I have a rough estimate of a figure I am aiming for.

The question says to use the standard integral: .$\displaystyle \int \frac{dx}{\sqrt{k^{2}-x^{2}}}\:=\: \tan^{-1}\!\left ( \frac{x}{\sqrt{k^{2}-x^{2}}} \right )+C$
. . This is NOT a standard formula!

I am struggling to interpret the $\displaystyle r+\frac{rd}{\sqrt{r^2-x^2}}$ into the form $\displaystyle \frac{dx}{\sqrt{k^2-x^2}}$ in order to evaluate the integral.
. . Really? . . . What's stopping you?

Any help or guidance would be appreciated as I have spent too much time trying to do this. Thanks.

That "standard integral" would be: .$\displaystyle \int\frac{dx}{\sqrt{r^2-x^2}} \:=\:\sin^{\text{-}1}\!\left(\frac{x}{r}\right)+ C$

We have: .$\displaystyle 2\pi r \int^{110}_{\text{-}70}\left(1 + \frac{d}{\sqrt{r^2-x^2}}\right)\,dx \;=\;2\pi r\left[x + d\sin^{-1}\!\left(\frac{x}{r}\right)\right]^{110}_{\text{-}70}$

Now finish it . . .

5. ## Re: Definite intergral

Thank you for your help, I have finished the problem now

I was just having a basic algebra problem, and that standard integral information being incorrect threw me a bit