# Thread: Integral Area

1. ## Integral Area

I don't really understand this question:
Evaluate:
$\int_0^3 x \sqrt{81-x^4}\, dx$

by making a substitution of the form $u=x^p$ and then interpreting the result in terms of a known area.

I get that I should find a usable $u=x^p$ then evaluate but don't understand what it means by interpreting in terms of a known area.

Thanks

2. ## Re: Integral Area

Originally Posted by Bowlbase
I don't really understand this question:
Evaluate:
$\int_0^3 x \sqrt{81-x^4}\, dx$

by making a substitution of the form $u=x^p$ and then interpreting the result in terms of a known area.

I get that I should find a usable $u=x^p$ then evaluate but don't understand what it means by interpreting in terms of a known area.

Thanks
$u = x^2$

$du = 2x \, dx$

$\frac{1}{2} \int_0^3 2x \sqrt{81-x^4} \, dx$

substitute and reset the limits of integration ...

$\frac{1}{2} \int_0^9 \sqrt{81-u^2} \, du$

now, what common geometric figure is formed by the graph of $y = \sqrt{81-u^2}$ ?

3. ## Re: Integral Area

semi-circle. So, this would the area y=0 to y=9?

4. ## Re: Integral Area

Originally Posted by Bowlbase
semi-circle. So, this would the area y=0 to y=9?
Surely it would be the area of a quarter of the circle 9 units in radius centred at the origin...

If you want to use integrals, then make the substitution $\displaystyle u = \sin{\theta} \implies du = \cos{\theta}\,d\theta$...

5. ## Re: Integral Area

Originally Posted by Prove It
Surely it would be the area of a quarter of the circle 9 units in radius centred at the origin...
don't forget the $\frac{1}{2}$ in front of the definite integral ...

6. ## Re: Integral Area

Okay, I think I understand. If I take 1/8 of pir^2 I should have the correct answer then, no?

7. ## Re: Integral Area

Originally Posted by Bowlbase
Okay, I think I understand. If I take 1/8 of pir^2 I should have the correct answer then, no?
Yes. You might also want to try solving the integral directly as well, just to verify your answer

8. ## Re: Integral Area

Yes, of course. I doubt my instructor would accept it otherwise.

9. ## Re: Integral Area

Okay, I thought I had this sorted out last night but obviously didn't think too much as I worked this problem.

First:
Having gotten $\frac{1}{2} \int_0^9 \sqrt{81-u^2} \, du$
It's not clear to me why we're changing the range from 3 to 9. From a geometric point of view it makes perfect sense but just taking it in the context of the integral I don't see the reasoning.

Second: I thought that it would be simple to evaluate this at first but we have not done anything with squares inside roots. Do I need to make another substitution for this? For instance $c=u^2$?

Thanks for the help so far!

10. ## Re: Integral Area

When you change the integral from an integral with respect to x to an integral with respect to u, the terminals need to change to u values.

Since \displaystyle \begin{align*} u = x^2 \end{align*}, when \displaystyle \begin{align*} x = 0, u = 0 \end{align*} and when \displaystyle \begin{align*} x = 3, u = 9 \end{align*}.

Now, as for solving the integral, like I suggested, you need to make the substitution \displaystyle \begin{align*} u = 9\sin{\theta} \implies du = 9\cos{\theta}\,d\theta \end{align*}, and note that when \displaystyle \begin{align*} u = 0, \theta = 0\end{align*} and when \displaystyle \begin{align*} u = 9, \theta = \frac{\pi}{2} \end{align*} and the integral becomes...

\displaystyle \begin{align*} \frac{1}{2}\int_0^9{ \sqrt{ 81 - u^2 } \,du } &= \frac{1}{2}\int_0^{ \frac{\pi}{2} }{ \sqrt{ 81 - \left( 9\sin{\theta} \right)^2 }\,9\cos{\theta} \, d\theta } \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}} { \sqrt{ 81 - 81 \sin^2{\theta} } \,\cos{\theta}\,d\theta} \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}} { \sqrt{ 81\left(1 - \sin^2{\theta}\right) } \,\cos{\theta}\,d\theta} \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}} { \sqrt{ 81\cos^2{\theta} } \,\cos{\theta}\,d\theta} \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}}{ 9\cos{\theta}\cos{\theta}\,d\theta} \\ &= \frac{81}{2}\int_0^{\frac{\pi}{2}} {\cos^2{\theta}\,d\theta} \\ &= \frac{81}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta} \\ &= \frac{81}{2}\left[\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} \right]_0^{\frac{\pi}{2}} \end{align*}

\displaystyle \begin{align*} &= \frac{81}{2}\left[\left(\frac{1}{2}\cdot \frac{\pi}{2} + \frac{1}{4}\sin{2\cdot \frac{\pi}{2}}\right) - \left(\frac{1}{2}\cdot 0 + \frac{1}{4}\sin{2\cdot 0}\right)\right] \\ &= \frac{81}{2} \left[\left( \frac{\pi}{4} + \frac{1}{4}\sin{\pi}\right) - \left(0 + \frac{1}{4}\sin{0}\right)\right] \\ &= \frac{81\pi}{8} \end{align*}

which is the same as half of the area of a quarter circle of radius 9 as discussed.

11. ## Re: Integral Area

We have yet to introduce a trig function into a integral, or derivative for that matter, that did not already have one in my class. So I never would have even guessed to do this...

Are there rules that precede doing this? Are we able to do this because it is part of a circular function?

12. ## Re: Integral Area

Originally Posted by Bowlbase
We have yet to introduce a trig function into a integral, or derivative for that matter, that did not already have one in my class. So I never would have even guessed to do this...

Are there rules that precede doing this? Are we able to do this because it is part of a circular function?
I suggest reading this to learn about Trigonometric Substitution.

13. ## Re: Integral Area

Originally Posted by Prove It
I suggest reading this to learn about Trigonometric Substitution.

This helped out a lot, thank you.