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