lfdahl said:
Let $R$ be the region $\left\{(x, y) : 0 \leq x \leq 1, 3^x − x − 1 \leq y \leq x\right\}$.
Find the volume of the solid obtained by rotating $R$ around the line $y = x$.
[sp]Rotate the axes through $45^\circ$ by introducing new coordinates $u = \dfrac{x+y}{\sqrt2}$, $v = \dfrac{y-x}{\sqrt2}$. The line $y=x$ then becomes the $u$-axis.
If $y = 3^x − x − 1$ then $x+y+1 = 3^x$, and so $\ln(x+y+1) = x\ln3$. In terms of $u$ and $v$ that becomes $\ln(\sqrt2u + 1) = \dfrac{\ln3}{\sqrt2}(u-v)$. Solve that for $v$ to get $$v = u - \frac{\sqrt2}{\ln3}\ln(\sqrt2u + 1).$$
The volume of the solid obtained by rotating $R$ around the $u$-axis is then $$V = \pi\!\!\int_0^{\sqrt2}v^2\,du = \pi\!\!\int_0^{\sqrt2}\left(u - \frac{\sqrt2}{\ln3}\ln(\sqrt2u + 1)\right)^{\!\!2}du.$$ Substitute $t = \sqrt2u+1$: $$V = \pi\!\!\int_1^3\left(\frac{t-1}{\sqrt2} - \frac{\sqrt2}{\ln3}\ln t\right)^{\!\!2}\frac{dt}{\sqrt2} = \frac\pi{\sqrt2}\int_1^3\left(\frac12(t-1)^2 - \frac2{\ln3}(t-1)\ln t + \frac2{(\ln3)^2}(\ln t)^2\right)dt.$$
Dealing with the three terms in that last integral, the first one is easy, the other two require integration by parts: $$\int_1^3 (t-1)^2dt = \Bigl[\frac13(t-1)^3\Bigr]_1^3 = \frac83;$$ $$ \int_1^3 (t-1)\ln t\,dt = \Bigl[\frac12(t-1)^2\ln t \Bigr]_1^3 - \int_1^3 \frac12(t^2 - 2t + 1)\frac1t\,dt = 2\ln3 - \Bigl[\frac14t^2 - t + \frac12\ln t\Bigr]_1^3 = 2\ln 3 - \Bigl(-\frac34 + \frac12\ln3 + \frac34\Bigr) = \frac32\ln3; $$ $$ \int_1^3(\ln t)^2dt = \Bigl[t(\ln t)^2 \Bigr]_1^3 - \int_1^3t\frac{2\ln t}tdt = 3(\ln3)^2 - 2\Bigl[t\ln t - t\Bigr]_1^3 = 3(\ln3)^2 - 6\ln3 + 4.$$
Put those results into the integral for $V$ to get $$V = \frac\pi{\sqrt2}\left(\frac43 - 3 + 6 - \frac{12}{\ln3} + \frac8{(\ln3)^2}\right) = \frac\pi{\sqrt2}\left(\frac{13}3 - \frac{12}{\ln3} + \frac8{(\ln3)^2}\right).$$
Numerically, that gives $V\approx 0.086$, which looks about the right order of magnitude from the appearance of a graph of the function.
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