Volume of Revolution of area included between Sine and Exp curves

Question

Question: Find the volume of the solid generated by rotating the region bounded by the curves below about the horizontal line $y=-2$.

$x= \frac{3\pi}{2}$, $y=e^{x-3}+2$

$x = 0$, $y=\sin x$

Here is what the graph look like.

I used washer.

\begin{align} R_\mathrm{big} &= e^{x-3} + 2 + 2 \\ R_\mathrm{big} &= e^{x-3} + 4\\ R_\mathrm{small} &= \sin x+ 2 \\ V &= \pi \int_0^{\frac{3\pi}{2}} \left[ \left(e^{x-3} + 4\right)^2 - \left(\sin x+ 2 \right)^2 \right]\,\mathrm{d}x \stackrel{\text{calc}}{\implies} 343.9678\,\mathrm{units}^3. \end{align}

I was wondering if my work is correct.

Thanks

Answer 1

Your integral is set up correctly. And

$$\begin{split}\pi\int_0^{3\pi/2}\left((e^{x-3}+4)^2-(\sin x +2)^2\right)dx&=\pi\int_0^{3\pi/2}\left(e^{2x-6}+8e^{x-3}+16-(\underbrace{\sin^2x}_{=\frac 1 2(1-\cos 2x)}+4\sin x+4)\right)dx\\ &=\pi\int_0^{3\pi/2}\left(e^{2x-6}+8e^{x-3}+\frac {23}{2} +\frac 1 2\cos2x-4\sin x\right)dx\\ &=\pi\left[\frac 1 2e^{2x-6}+8e^{x-3}+\frac {23} 2x+\frac 1 4\sin 2x+4\cos x\right]_{0}^{3\pi/2}\\ &=\pi\left[\left(\frac 1 2e^{3\pi-6}+8e^{3\pi/2-3}+\frac{69}{4}\pi+0+0\right)-\left(\frac 1 2e^{-6}+8e^{-3}+0+0+4\right)\right]\\ &\approx343.97\end{split}$$

It looks to be correct.

Answer 2

Trying to guess at your question:

Find volume by sweeping area between curves

$$y_1= e^{x-3}+2,\; y_2= \sin x $$

about a line of symmetry parallel to x-axis

$$y=-2$$

between x-limits $( 0, \frac{3 \pi}{2})$ by washer method.

Your integral setting up and result then seem to be alright.