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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ahat on July 08, 2013, 04:33:39 pm

Title: Volumes of solids of revolution?
Post by: ahat on July 08, 2013, 04:33:39 pm: Find the volume of the solid of revolution formed when the region between the y-axis and the curves of y =cos^-1(x) and y = sin^-1(x) is rotated about the y-axis.

I don' think I'm approaching it correctly. I did V = $\pi$ ∫ cos²(y) - sin²(y) dy between 0 and $\frac{\pi}{2}$ but obviously this equals 0. I don't think I'm interpreting the question correctly.

Help appreciated.
Title: Re: Volumes of solids of revolution?
Post by: b^3 on July 08, 2013, 05:06:23 pm: Do a quick sketch first, here's one I prepared earlier :P
(https://s3.amazonaws.com/grapher/exports/xpevueiky9.png)
Now we are rotating around the $y$ axis, which in this case means you need to split the integral into two regions, adding the volumes formed by rotating the red curve from $0$ to the intersection of the curves (remember since we have $dy$ this will be the corresponding $y$ coordinate) and then rotated the blue curve from there to $y=\frac{\pi}{2}$ . This means you will need to find the point of intersection of the two curves (check your value for that, and then remember to take the $y$ value).

Spoiler
$\begin{alignedat}{1}V & =\pi\left(\int_{0}^{\frac{\pi}{4}}\sin^{2}\left(y\right)dy+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos^{2}\left(y\right)dy\right)\end{alignedat}$

Hope that helps :)
Title: Re: Volumes of solids of revolution?
Post by: ahat on July 08, 2013, 05:37:50 pm: Thanks heaps b^3 :) I'm still a bit confused as to why the integral needs to be split? Why is it that the
V = π ∫ [f(y)]² - [g(y)]² dy doesn't hold in this case? Apologies for my ignorance.
Title: Re: Volumes of solids of revolution?
Post by: b^3 on July 08, 2013, 05:46:55 pm: That rule would work if the region we wanted was between the two curves, $f(x)$ and $g(x)$ , but in our case the region is not between those curves, rather it's between either of those curves and the $y$ axis. In other words we are rotating the shaded region below around the $y$ axis.
(http://content.screencast.com/users/tbatty/folders/Jing/media/41a63234-d500-429f-83aa-c61d2573d3b3/2013-07-08_1750.png)
Now we have the region bounded by a different curve depending on the $y$ value. For $0\leq x\leq \frac{\pi}{4}$ it's bounded by the red curve and the $y$ axes while for $\frac{\pi}{4}\leq x \leq \frac{\pi}{2}$ it's bounded by the blue curve and the $y$ axis. So we can split this region up into two regions, and rotate them separately, and then add the volumes of those two regions.
Title: Re: Volumes of solids of revolution?
Post by: ahat on July 08, 2013, 06:47:33 pm: Oh my gosh b^3 hahaha. So simple, why did I not get that. Thanks so much, +1 respect,
Title: Re: Volumes of solids of revolution?
Post by: lzxnl on July 08, 2013, 07:01:55 pm: With EVERY SINGLE area or volume question, draw a rough sketch of just exactly what you're finding first. Trust me, it'll paradoxically save you time and marks lost.