Author Topic: Question help? (Read 1101 times) Tweet Share

Yoda · « **on:** May 20, 2014, 06:44:47 pm »

Find the volume between the lines x=0 and x=1 when the graph of y=tan^-1(x) is rotated around the y-axis.

b^3 · « **Reply #1 on:** May 20, 2014, 07:00:31 pm »

Misread the question, see below.

Yoda · « **Reply #2 on:** May 20, 2014, 07:09:25 pm »

The answer is pi(pi/2 -1)units^3 you found the volume about the y-axis but the graph is shaded between the lines x=0 and x=1. I'm a bit stuck on this one.

b^3 · « **Reply #3 on:** May 20, 2014, 07:16:43 pm »

Quote from: Yoda on May 20, 2014, 07:09:25 pm

The answer is pi(pi/2 -1)units^3 you found the volume about the y-axis but the graph is shaded between the lines x=0 and x=1. I'm a bit stuck on this one.

Whoops, my bad, misread the question (this is why you should always do a little sketch before doing vol of revs questions!) In that case you have $1-\tan^2(x)$ instead and integrate that. It's the same method with a few adjustments. That's because we're rotating the volume "below" (to the left to the y axis) x=1 and minusing the volume "below" y=tan^-1(x).

$\begin{alignedat}{1}V & =\pi\int_{y=a}^{y=b}x^{2}dy \\ y & =\tan^{-1}\left(x\right) \\ x & =\tan\left(y\right) \\ \implies x^{2} & =\tan^{2}\left(y\right) \\ x=0, & \: y=\tan^{-1}\left(0\right) \\ & y=0 \\ x=1,\: & y=\tan^{-1}\left(1\right) \\ & y=\frac{\pi}{4} \\ \sin^{2}\left(x\right)+\cos^{2}\left(x\right) & =1 \\ \tan^{2}\left(x\right)+1 & =\sec^{2}\left(x\right) \\ \tan^{2}\left(x\right) & =\sec^{2}\left(x\right)-1 \\ V & =\pi\int_{0}^{\frac{\pi}{4}}\left(1-\tan^{2}\left(y\right)\right)dy \\ & =\pi\int_{0}^{\frac{\pi}{4}}\left(1-\sec^{2}\left(x\right)+1\right)dy \\ & =\pi\left(\int_{0}^{\frac{\pi}{4}}\left(2\right)dy-\int_{0}^{\frac{\pi}{4}}\left(\sec^{2}\left(x\right)\right)dy\right) \\ \text{As }\frac{d}{dx}\left(\tan\left(x\right)\right)=\sec^{2}\left(x\right) \\ \implies\tan\left(x\right)=\int\sec^{2}\left(x\right)dx \\ V & =\pi\left(\left[2y\right]_{0}^{\frac{\pi}{4}}-\left[\tan\left(x\right)\right]_{0}^{\frac{\pi}{4}}\right) \\ & =\pi\left(2\left(\frac{\pi}{4}\right)-2\left(0\right)-\tan\left(\frac{\pi}{4}\right)+\tan\left(0\right)\right) \\ \therefore V & =\pi\left(\frac{\pi}{2}-1\right)\:\text{units}^{3} \end{alignedat}$

If you wanted to work it back from the answer from before you could just take that away from the cylinder formed from rotating $x=1$ .
$\begin{alignedat}{1}V & =V_{2}-V_{1} \\ & =\pi r^{2}h-\pi\left(1-\frac{\pi}{4}\right) \\ & =\pi\left(1\right)^{2}\tan^{-1}\left(1\right)-\pi\left(1-\frac{\pi}{4}\right) \\ & =\pi\cdot\frac{\pi}{4}-\pi\left(1-\frac{\pi}{4}\right) \\ & =\frac{\pi^{2}}{4}-\pi+\frac{\pi^{2}}{4} \\ & =\pi\left(\frac{\pi}{2}-1\right) \end{alignedat}$

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Yoda

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