부분적분법

For a fixed value of $x$, if we rotate about the $x$-axis, we form a washer of inner radius $B(x)$ and outer radius $T(x)$ and hence of area $\pi [T(x)^2 - B(x)^2]$. We integrate this function from $x=0$ to $x=3$ to find the total volume $V$:

$$\begin{aligned} V &= \int_0^3 \pi [T(x)^2 - B(x)^2]\,\mathrm{d}x \\ &= \pi \int_0^3 (\sqrt{x}e^{3x})^2 - (\sqrt{x}(1+2x))^2 \,\mathrm{d}x \\ &= \pi \int_0^3 \big( xe^{6x} - (x+4x^2+4x^3) \big) \,\mathrm{d}x \\ &= \pi \int_0^3 xe^{6x} \,\mathrm{d}x - \pi \Big[ \frac{x^2}{2} + \frac{4x^3}{3} + x^4\Big]_{0}^3 \\ &= \pi \int_0^3 xe^{6x} \,\mathrm{d}x - \pi \Big[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4\Big] \end{aligned}$$

For the first integral, we use integration by parts with $u(x) = x$, $\mathrm{d}v = e^{6x}\mathrm{d}x$, so that $\mathrm{d}u=\mathrm{d}x$ and $v(x)=\frac16e^{6x}$:

$$\begin{aligned} \int_0^3 xe^{6x} \,\mathrm{d}x &=\underbrace{ \frac{xe^{6x}}{6}}_{uv}\bigg|_{0}^3 - \int_0^3 \underbrace{\frac{1}{6}e^{6x} \,\mathrm{d}x }_{v\mathrm{d}u}\\ &= \frac{3e^{18}}{6} - 0 - \frac{1}{36} e^{6x} \bigg|_{0}^3 = \frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg). \end{aligned}$$

Therefore, the total volume is

$$\begin{aligned} V = \pi \bigg[\frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg) \bigg] - \pi \bigg[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4 \bigg] = \pi \bigg( \frac{17 e^{18}-4373}{36} \bigg). \end{aligned}$$