Areas between curves

**Fnus** · October 17th 2008, 11:12 PM

So, I've been trying to do this on my own, but I'm really at a loss.
(The problem is in the attatched file)

At first I figured I'd need to find out where the two curves intersect,
so I did $x = \sqrt{k}$

Which I dont even know if right. I sure hope it is!

Then I figured that the area would go from 0 to $\sqrt{k}$ ,
and then I tried to integrate, and substitute x for $\sqrt{k}$ .

Eventually I ended up with something like 1.8-something which is wrong.

I'd love for some guidance on this one, thanks in advance!

**Chop Suey** · October 17th 2008, 11:45 PM

The graph does intersect at $x = \sqrt{k}$

Therefore, the integral is:
$\int_{0}^{\sqrt{k}} k-x^2~dx$

Simply integrate it, and find the area between sqrt(k) and 0. Then set it equal to 2.4 and solve for k. You should be getting k = 2.3489...

$k \sqrt{k} - \frac{\sqrt{k^3}}{3} = 2.4$

$\frac{2\sqrt{k^3}}{3} = 2.4$

k = ...

Your approach is correct, perhaps you might have made a mistake in solving for k or while integrating.

**Fnus** · October 18th 2008, 06:49 AM

Okay, thankies.

At least I was on the right track, Imma just keep doing it 'til I get it right, then.
I was just afraid I was on the complete wrong track, thanks for clearing it up. (:

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