Q. 12 - Calculus

Question

Set up and solve definite integrals to find each volume, surface area, or arc length that follows. Solve each volume problem both with disks/washers and with shells, if possible.

The arc length of the curve is traced out by the graph of $f\left(x\right) = \ln (csc x)$ on the interval $\left[ \frac{\pi }{4} , \frac{\pi }{2} \right]$.

Step-by-Step Solution

Answer

$\ln \left(\sqrt{2}+1\right)$.

Step 1. Given Information.

The volume of a given solid 

$f\left(x\right)=\ln \left(cscx\right)$.

Step 2. Formulation.

The length of arc of a curve is 

${\int }_a^b\sqrt{1+y{\text{'}}^2}dx$.

So, The arc is 

$$\begin{gathered} y=\ln \left(cscx\right) \\ y\text{'}=\frac{1}{cscx}·\left(-cscx cotx\right) =-cotx \\ L={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\sqrt{1+{(-cotx)}^2} dx \\ L= {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\sqrt{1+{\left(cotx\right)}^2} dx [1+co{t}^{2}x=cs{c}^{2}x] \\ L= {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\sqrt{cs{c}^{2}x} dx \\ L= {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}cscx dx \\ L= {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{cscx \left(cscx+cotx\right)}{\left(cscx+cotx\right)} dx \\ L= {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{cs{c}^{2}x +\left(cscxcotx\right)}{\left(cscx+cotx\right)} dx \\ we know that cscx+cotx =u \\ \Rightarrow -cscx cotx -cs{c}^{2}x dx=du \\ L=-{\left[\ln \left|cscx+cotx\right|\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ L=-\ln \left(1+0\right)+\ln \left(\sqrt{2}+1\right) \\ L=\ln \left(\sqrt{2}+1\right) \end{gathered}$$