Find the volume between $y = 4 − \frac{3x}{2}$ and $y=0$ and $x\in [0, 1]$

Question

Find the volume $V$ of the solid obtained by rotating the region bounded by the given curves about the specified line.

$$y = 4 − \dfrac{3}{2x},\ y = 0,\ x = 0,\ x = 1$$

about the $x$-axis.

I keep getting $\dfrac{61\pi}{4}$ and its wrong. HELP! Any tips would be wonderful!!

Answer 1

If I have not misunderstood your given curve,

$$\begin{align} V =&\int_0^1 \pi\left(4-\frac{3x}{2}\right)^2dx\\ =&\pi\int_0^1 \left(16-12x+\frac{9x^2}{4}\right)dx\\ =&\pi\left[16x-6x^2+\frac{3x^3}{4}\right]_0^1\\ =&\pi\left(16-6+\frac{3}{4}\right)\\ =&\frac{43\pi}{4}\\ \end{align}$$

Answer 2

\begin{align} V &= \int_{3/8}^{1}\pi\left(4 - {3 \over 2x}\right)^{2}\,{\rm d}x = \pi\int_{3/8}^{1}\left(16 - {12 \over x} + {9 \over 4x^{2}}\right)\,{\rm d}x = \pi\left.\vphantom{\Huge A}% \left\{16x - 12\ln\left(x\right) - {9 \over 4x}\right\}\right\vert_{3/8}^{1} \\&= \pi\left[% 16 - {9 \over 4} - 6 + 12\ln\left(3 \over 8\right) + 6 \right] = \color{#ff0000}{\large% \left[{55 \over 4} + 12\ln\left(3 \over 8\right)\right]\pi} \end{align}