Applications of Integration: Vol.: \(y=x\ln x\), \(x\)-axis — ACJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
The region \(R\) is bounded by the curve with equation \(y = x\ln x\), the lines \(x=\mathrm{e}\), \(x=\mathrm{e}^2\) and the \(x\)-axis. Find the exact volume of the solid formed when \(R\) is rotated about the \(x\)-axis by \(2\pi\) radians. Give your answer in the form \(\dfrac{\pi \mathrm{e}^3}{27}\!\left(a\mathrm{e}^3+b\right)\), where \(a\) and \(b\) are integers to be found.
Show full worked solution▾
Using integration by parts twice. First, let \(I = \displaystyle\int x^2(\ln x)^2\,\,\mathrm{d} x\): \[\begin{aligned} I &= \frac{x^3}{3}(\ln x)^2 - \int\frac{x^3}{3}\cdot\frac{2\ln x}{x}\,\,\mathrm{d} x\\ &= \frac{x^3}{3}(\ln x)^2 - \frac{2}{3}\int x^2\ln x\,\,\mathrm{d} x \end{aligned}\]
For \(\displaystyle\int x^2\ln x\,\,\mathrm{d} x\), integrate by parts again: \[\begin{aligned} \int x^2\ln x\,\,\mathrm{d} x &= \frac{x^3}{3}\ln x - \int\frac{x^3}{3}\cdot\frac{1}{x}\,\,\mathrm{d} x\\ &= \frac{x^3}{3}\ln x - \frac{x^3}{9} + C \end{aligned}\]
Therefore: \[\begin{aligned} I &= \frac{x^3}{3}(\ln x)^2 - \frac{2}{3}\!\left(\frac{x^3}{3}\ln x - \frac{x^3}{9}\right) + C\\ &= \frac{1}{27}x^3\!\left(9(\ln x)^2 - 6\ln x + 2\right) + C \end{aligned}\]
Evaluating: \[\begin{aligned} \pi\int_{\mathrm{e}}^{\mathrm{e}^2} x^2(\ln x)^2\,\,\mathrm{d} x &= \frac{\pi}{27}\Bigl[x^3\!\left(9(\ln x)^2-6\ln x+2\right)\Bigr]_{\mathrm{e}}^{\mathrm{e}^2}\\ &= \frac{\pi}{27}\Bigl(\mathrm{e}^6(9\cdot4 - 12 + 2) - \mathrm{e}^3(9\cdot1 - 6 + 2)\Bigr)\\ &= \frac{\pi}{27}\!\left(26\mathrm{e}^6 - 5\mathrm{e}^3\right)\\ &= \frac{\pi\mathrm{e}^3}{27}\!\left(26\mathrm{e}^3 - 5\right) \end{aligned}\]