Find the volume $V$ as a result of revolving the region between $f(x) = x^2 - 2x$ and $g(x) = x$ around the line $y = 4$.

Question

For context, I'm self-studying calculus and am doing some practice problems to test my understanding of solids of revolution. I want to ask:

Are my solutions (using both Disc and Shell Integration) to the following question correct and efficient? (As in not overcomplicating things.) Also, is each step justified?

Find the volume $V$ as a result of revolving the region between $f(x) = x^2 - 2x$ and $g(x) = x$ around the line $y = 4$.

I used Desmos to help me visualize the problem:

(This will particularly be helpful for the Shell Method solution).

• Disc Method (or specifically the Washer Method for hollow solids of revolution): Integrating parallel to the axis of revolution

$$\begin{align*} V &= \int_a^b \pi\, (4 - f(x))^2 - \pi\, (4 - g(x))^2 \,dx \tag{$\scriptsize f(x)=g(x) \iff x^2 - 2x = x$} \\ &= \pi \int_0^3 (4 - x^2 + 2x)^2 - (4 - x)^2 \,dx \tag{$\scriptsize x(x - 3) = 0 \implies x = 0 = a \lor x = 3 = b$} \\ &= \pi \int_0^3 (x^4 - 4x^3 - 5x^2 + 24x) \,dx\\ & = \pi \,\left.\left(\frac{x^5}{5} - x^4 - \frac{5x^3}{3} + 12x^2 \right)\right|_0^3 \\ & = \pi \left(\frac{243}{5} - 81 - \frac{135}{3} + 108\right)\\ & = \pi\left(\frac{243 - 405 -225 + 540}{5}\right)\!, \\[7pt] V &= \frac{153\pi}{5} \text{ units}^3. \;\llap{\mathrel{\bbox[5px, border: 2px solid red]{\phantom{V = \frac{153\pi}{5} \text{ units}^3.}}}} \end{align*}$$

• Shell Method: Integrating perpendicular to the axis of revolution

$$\begin{align*} f^{-1}(y) &= \begin{cases} 1 - \sqrt{y + 1}, &\text{if $\,x \leq 1$} \\ 1 + \sqrt{y + 1}, &\text{if $\,x \geq 1$.} \end{cases} \\[5pt] g^{-1}(y) &= x. \\[5pt] V &= \overbrace{\color{red}{\text{Revolved Red Region}}}^{\color{red}{R}} \,+\, \overbrace{\color{green}{\text{Revolved Green Region}}}^{\color{green}{G}} \,+\, \overbrace{\color{blue}{\text{Revolved Blue Region}}}^{\color{blue}{B}} \\[5pt] &= \underbrace{\color{red}{\int_{-1}^0 2\pi (4 - y) (1 - [1 - \sqrt{y + 1}]) \,dy}}_{\color{red}{R}} \,+ \underbrace{\color{green}{\int_{-1}^0 2\pi (4 - y) ([1 + \sqrt{y + 1}] - 1) \,dy}}_{\color{green}{G}} \,+ \\ &\phantom{=}\; \underbrace{\color{blue}{\int_0^3 2\pi (4 - y) ([1 + \sqrt{y + 1}] - y) \,dy}}_{\color{blue}{B}}. \\[5pt] \frac{V}{2\pi} &= \color{red}{\int_{-1}^0 (4 - y) \sqrt{y + 1} \;dy} \,+ \color{green}{\int_{-1}^0 (4 - y) \sqrt{y + 1} \;dy} \,+ \color{blue}{\int_0^3 (4 - y) \,dy} \,+ \\ &\phantom{=}\; \color{blue}{\int_0^3 (4 - y) \sqrt{y + 1} \;dy}\,+ \color{blue}{\int_0^3 (y^2 - 4y) \,dy} \\[5pt] &= \underbrace{\color{purple}{\int_{-1}^3 (4 - y) \sqrt{y + 1} \;dy}}_{\text{Let } u \,=\, y \,+\, 1 \implies du \,=\, dy} \,+ \underbrace{\color{green}{\int_{-1}^0 (4 - y) \sqrt{y + 1} \;dy}}_{\text{Let } u \,=\, y \,+\, 1 \implies du \,=\, dy} \,+ \color{blue}{\int_0^3 (y^2 - 5y + 4) \,dy} \\[5pt] &= \color{purple}{\int_{u=0}^{u=4} (5 - u) \sqrt{u} \;du} \,+ \color{green}{\int_{u=0}^{u=1} (5 - u) \sqrt{u} \;du} \,+ \color{blue}{\frac{1}{3} (3)^3} - \color{blue}{\frac{5(3)^2}{2}} + \color{blue}{4(3)} \\[5pt] &= \color{purple}{\left(\frac{10}{3} (4) \sqrt{4} - \frac{2}{5} (4)^2 \sqrt{4}\right)} + \color{green}{\left(\frac{10}{3} (1) \sqrt{1} - \frac{2}{5} (1)^2 \sqrt{1}\right)} -\, \color{blue}{\frac{3}{2}} \\[5pt] &= \color{purple}{\frac{208}{15}} + \color{green}{\frac{44}{15}} - \color{blue}{\frac{3}{2}} = \frac{416 + 88 - 45}{30} = \frac{459}{30} = \frac{153}{10}, \\[5pt] V &= \frac{153}{10} \cdot 2\pi, \\[7pt] V &= \frac{153\pi}{5} \text{ units}^3. \;\llap{\mathrel{\bbox[5px, border: 2px solid red]{\phantom{V = \frac{153\pi}{5} \text{ units}^3.}}}} \end{align*}$$