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Manipulating equations.

shamieh

Active member
Sep 13, 2013
539
I know this is easy but im getting hung up on this problem for some reason.

Suppose we have \(\displaystyle \sqrt{8}sin\theta\) and we plug in \(\displaystyle 2\) for \(\displaystyle \theta\).

\(\displaystyle \sqrt{8}sin(2).\)

how are they then coming to the conclusion of \(\displaystyle \sin\theta = \frac{1}{\sqrt{2}}\)

That makes NO sense! We just plugged a \(\displaystyle 2\) in for \(\displaystyle \theta\)! Where does the extra \(\displaystyle \theta \) come from?!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
I'm afraid you're going to have to give us some context for this question. Can you please give us the original problem statement, word-for-word?

(As a side note, you can't go from an expression to an equation, so yes, I'm very puzzled by that transition myself.)
 

shamieh

Active member
Sep 13, 2013
539
My teacher has left out some weird step that I am missing! Originally I am solving this \(\displaystyle 2\) \(\displaystyle \int^2_0 \sqrt{8 - x^2} - \frac{1}{2}x^2 dx\)

Ignore the second, that is easy so I split it up into two different integrals. Here is what I have so far, notice I am only doing the first one.

\(\displaystyle 2 \int^2_0 \sqrt{8 -x^2} dx\)

\(\displaystyle 2 \int^2_0 \sqrt{8 - \sin^2\theta} * \sqrt{8}\cos\theta d\theta\)

Also note that I have: \(\displaystyle 1 - \sin^2\theta = cos^2\theta\)
\(\displaystyle 8 - 8\sin^2\theta = 8\cos^2\theta\)
\(\displaystyle x^2 = 8sin^2\theta\)
\(\displaystyle x = \sqrt{8}\sin\theta\)
\(\displaystyle dx = \sqrt{8}\cos\theta \, d\theta\)

So now since I've done the substitution i need to update the limits... So my teacher said ok plug in the old limits into x.. Which I understand the first limit. He plugs in a 0. \(\displaystyle \sqrt{8}sin(0) \)= 0 thus the lower limit is 0 but for the upper limit he plugs in a 2 ....\(\displaystyle \sqrt{8}sin(2)\) and somehow gets \(\displaystyle 2 = \sqrt{8}sin\theta\) which becomes
\(\displaystyle sin\theta = \frac{1}{\sqrt{2}} \)

Like what am I missing here??
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Ah, thanks much for that background. Let's consider this integral:
$$\int_{0}^{2} \sqrt{8-x^{2}} \, dx.$$
I would indeed do a trig substitution. I think you have a bit of ambiguity in your substitution. Draw a right triangle, angle $\theta$, hypotenuse $\sqrt{8}$, and opposite side $x$. Then the adjacent side is $\sqrt{8-x^{2}}$. You have that
\begin{align*}
\frac{x}{ \sqrt{8}}&= \sin(\theta) \\
\frac{dx}{ \sqrt{8}}&= \cos(\theta) \, d\theta \\
\frac{ \sqrt{8-x^{2}}}{ \sqrt{8}}&= \cos( \theta)
\end{align*}
That will enable you do substitute in for the integrand and the differential. Now, the limits: $0$ and $2$ are $x$ values, because that is the variable in the differential. To get to $\theta$ values, you have to solve $0= \sqrt{8} \sin(\theta)$ for $\theta$, as well as $2= \sqrt{8} \sin( \theta)$ for $\theta$. There's no plugging $0$ or $2$ into the trig functions, because that's a category mistake: $0$ and $2$ are $x$-values, not $\theta$-values. Does that answer your question?
 

shamieh

Active member
Sep 13, 2013
539
Thanks so much!
 

shamieh

Active member
Sep 13, 2013
539
So is my work correct?


\(\displaystyle 2 \int ^\frac{\pi}{4}_0\)

\(\displaystyle 2 \int ^\frac{\pi}{4}_0 \sqrt{8} cos\theta * \sqrt{8}cos\theta\)


\(\displaystyle 2 \int ^\frac{\pi}{4}_0 8 cos^2\theta\)

then I brought the 8 out and got


\(\displaystyle 16 [ \theta - \frac{1}{2} sin2\theta] |^{\pi/4}_0\)
 

shamieh

Active member
Sep 13, 2013
539
I don;'t understand what ive done wrong. How are they getting \(\displaystyle 2\pi + \frac{4}{3} \) as the final answer?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Yes, I think your re-worked integral is correct. WolframAlpha confirms your result, both the original and the re-worked integral.