Science:Math Exam Resources/Courses/MATH103/April 2011/Question 04 (b)/Solution 1

Disc Method. In this solution, we consider the volume as a sum of slices, cut orthogonal to the $y$-axis. Each slice has volume $\pi r^{2}\Delta y$, where $r$ is its radius and $\Delta y$ is its thickness. The bowl will be made by rotating the shaded blue region in the 2D figure

xy 2D image

. If the slice is at height $y$, then its radius is equal to $x$, where $y=e^{x}$. Hence, $r=\ln y.$ In the limit $\Delta y\to 0,$

${\begin{aligned}V_{2}=\int _{1}^{e}\pi (\ln y)^{2}\,dy\end{aligned}}$

Note that the limits of integration are $f(0)=e^{0}=1$ and $f(1)=e^{1}=e,$ since $x$ ranges from 0 to 1.

We use integration by parts to evaluate this integral, letting $f(y)=g(y)=\ln y.$ So in order to perform the integration by parts we need to recall the antiderivative of $\ln y$. To calculate this antiderivative we again use integration by parts:

${\begin{aligned}\int \ln ydy&=\int 1\cdot \ln y\,dy=y\ln y-\int y\cdot {\frac {1}{y}}\,dy\\&=y\ln y-\int 1\,dy=y\ln y-y+C.\end{aligned}}$

We now use this antiderivative to perform the integration by parts in the integral for $V_{2}.$ Hence,

${\begin{aligned}V_{2}&=\pi \int _{1}^{e}(\ln y)^{2}\,dy\\&=\pi \left[\ln y(y\ln y-y)\right]_{1}^{e}-\pi \int _{1}^{e}{\frac {1}{y}}(y\ln y-y)\,dy\\&=\pi \left[1(e\cdot 1-e)-0\right]-\pi \int _{1}^{e}(\ln y-1)\,dy\\&=-\pi \left[y\ln y-y-y\right]_{1}^{e}\\&=-\pi \left[e\ln e-e-e-1\ln 1+1+1\right]\\&=-\pi \left(-e+2\right)\\&=\pi \left(e-2\right).\\\end{aligned}}$

Reality check: Note that we obtain the same value for $V_{2}$ with both methods. To visualize the full bowl we can look at the 3D figure

Full 3D image from rotation

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