calculate the volume formed by rotation of a given region I have a question in elementary differential calculus:
Let $S$ be a region given by
$$S=\{(x,y): 0\leq x\leq 1,\ \ 3^x-x-1\leq y\leq x\}$$
then define $V$ the solid obtained by rotating $S$ around $$y=x$$, how to calculate the volume of $V$? Thanks for the help! 
 A: Rotate the coordinate system 45 degrees counter-clockwise to obtain a function that can be rotated about the $x$ axis.
This can be done by applying the transformation matrix
$$\begin{bmatrix}
\cos\left(-\frac{\pi}4\right) & -\sin\left(-\frac{\pi}4\right)\\
\sin\left(-\frac{\pi}4\right) & \cos\left(-\frac{\pi}4\right)
\end{bmatrix} = \frac1{\sqrt{2}}
\begin{bmatrix}
1 & 1\\
-1 & 1
\end{bmatrix}
$$
to a paramterization of $3^x-x-1$
$$\frac1{\sqrt{2}}
\begin{bmatrix}
1 & 1\\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
t\\
-3^t-t-1
\end{bmatrix} = 
\frac1{\sqrt{2}}\begin{bmatrix}
3^t-1\\
3^t-2t-1
\end{bmatrix}
$$
and converting to new coordinates by solving
$$\begin{bmatrix}
x\\
y
\end{bmatrix} = 
\frac1{\sqrt{2}}\begin{bmatrix}
3^t-1\\
3^t-2t-1
\end{bmatrix}$$
So, $x = \frac1{\sqrt{2}}\left(3^t-1\right) \implies t=\log_3\left(\sqrt{2}x+1\right)$
Plugging this into $3^t-2t-1$ gives the new function $$y=\frac1{\sqrt2}\left(3^{\log_3(\sqrt2x+1)} - 2\log_3(\sqrt2x+1)-1\right)=x-\sqrt2\log_3\left(\sqrt2x+1\right)$$
$y=x$ has now been transformed to $y=0$ and the integration limits $0$ and $1$ to $0$ and $\sqrt2$ respectively. 
Now calculate the volume using the normal method for rotations about the x axis.
$$V = \pi\int_0^{\sqrt2}y^2\;\text{d}x = \pi\int_0^{\sqrt2}\left(x-\sqrt2\log_3\left(\sqrt2x+1\right)\right)^2\;\text{d}x$$
which can be solved using standard integration methods. 
The answer according to WA is $\approx 0.0860724$.
