MHB Why is \sin\theta = \frac{1}{\sqrt{2}} when \theta = 2?

[shamieh]

I know this is easy but I am getting hung up on this problem for some reason.

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

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

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

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

 

[Ackbach]

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]

My teacher has left out some weird step that I am missing! Originally I am solving this $$2$$ $$\int^2_0 \sqrt{8 - x^2} - \frac{1}{2}x^2 dx$$

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.

$$2 \int^2_0 \sqrt{8 -x^2} dx$$

$$2 \int^2_0 \sqrt{8 - \sin^2\theta} * \sqrt{8}\cos\theta d\theta$$

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

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. $$\sqrt{8}sin(0) $$= 0 thus the lower limit is 0 but for the upper limit he plugs in a 2 ...$$\sqrt{8}sin(2)$$ and somehow gets $$2 = \sqrt{8}sin\theta$$ which becomes
$$sin\theta = \frac{1}{\sqrt{2}} $$

Like what am I missing here??

 

[Ackbach]

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]

Thanks so much!

 

[shamieh]

So is my work correct?$$2 \int ^\frac{\pi}{4}_0$$

$$2 \int ^\frac{\pi}{4}_0 \sqrt{8} cos\theta * \sqrt{8}cos\theta$$$$2 \int ^\frac{\pi}{4}_0 8 cos^2\theta$$

then I brought the 8 out and got$$16 [ \theta - \frac{1}{2} sin2\theta] |^{\pi/4}_0$$

 

[shamieh]

I don;'t understand what I've done wrong. How are they getting $$2\pi + \frac{4}{3} $$ as the final answer?

 

[Ackbach]

Yes, I think your re-worked integral is correct. WolframAlpha confirms your result, both the original and the re-worked integral.