# Integral Calculus, right or wrong?

I have two questions, which I decided and my feedback does not match that of my book, but do not seem to be wrong.

Calculate the volume of the solid obtained by rotating the region bounded by the curves

I) $y=e^x,\;y=0,\;x=0,\;x=1$ Around the $x$-axis

My answer was $$V=\pi\int_{0}^{1}\;\left(e^x\right)^2\;dx=\frac{\pi}{2}\int_0^1\;e^{2x}\cdot 2\;dx$$

$$\boxed{V=\frac{\pi}2\left(e^2-1\right)}$$

And my feedback says $$V=\frac{\pi}{12}\left( e^2-1 \right)$$



II) Have the other is $y=\sec x,\; y=1,\; x=-1,\;x=1$ Around the $x$-axis

$$V=\pi\int_{-1}^1\;\left( (\sec x)^2-1^2 \right)\;dx=\left.\pi\left( \tan x-x \right)\right|_{-1}^1$$

$$V=\pi(\tan(1)-2-\tan{(-1)})$$

And my feedback says $$V=2\pi(\tan1-1)$$



Am I wrong again?

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## 2 Answers

For the first one, I think someone's just accidentally written 12 instead of 2. I can't see any flaws in your working.

For the second, $$\tan(-1)=\frac{\sin(-1)}{\cos(-1)}=-\frac{\sin(1)}{\cos(1)}=-\tan(1)$$ $$\pi(\tan(1)−2−\tan(−1))=\pi(\tan(1)-2-(-\tan(1)))$$ $$=\pi(2\tan(1)-2)=2\pi(\tan(1)-1)$$

you were right but just needed to do a bit more :)

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Didactically perfect. –  marcelolpjunior Jun 26 at 23:03

In both cases you're correct. There is a typo in the feedback's answer in the first problem. To see why your answer is the same as the book's in the second case, recall that the tangent function is an odd function.

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