Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer\(x\)

\(y = \ln x,y = 1,y = 2,x = 0;\)about the \(y\)axis

Short answer

The volume of the solid is \(\frac{{\pi {e^2}}}{2}\left( {{e^2} - 1} \right){\rm{. }}\)

Step-by-step solution

definition of volume

Definition of Volume:

Consider a solid that lies between the curves\(x = a\), and\(x = b\). If the cross-section of\(S\)in the plane\({P_x}\), through and perpendicular to the\(x\)-axis, is given by an integrable function\(A(x)\), then the volume of\(S\)is

\(\begin{aligned}{l}V = \mathop {\lim }\limits_{\max \Delta {x_i} \to 0} \sum\limits_{i = 1}^n A \left( {x_i^*} \right)\Delta {x_i}\\V = \int_a^b A (x)dx\end{aligned}\)

solve for area and volume

It is given that the curves that bound a region are \(y = \ln x,y = 1,y = 2\), and \(x = 0\).

The specific line around which the region is rotated is\(y\)-axis.

Then, the equation\(y = \ln x\)is considered as\(x = {e^y}\).

If the curve\(x = {e^y}\)is rotated about the\(y\), a solid is obtained.

If is sliced at a point\(y\), a disk is obtained whose radius is given by\(x = {e^y}\).

Thus, the area of the disk is obtained as

\(\begin{aligned}{}A(y) = \pi {(x)^2}\\ = \pi {\left( {{e^y}} \right)^2}\\A = \pi {e^{2y}}\end{aligned}\)

The region lies between\(y = 1\), and\(y = 2\), so by definition the volume becomes

\(\begin{aligned}{}V &= \int_1^2 \pi \left( {{e^{2x}}} \right)dx\\ &= \pi \left( {\frac{{{e^{2x}}}}{2}} \right)_1^2\\V &= \pi \left( {\frac{{{e^{2(2)}}}}{2} - \frac{{{e^{2(1)}}}}{2}} \right)\end{aligned}\)

\( = \frac{{\pi {e^2}}}{2}\left( {{e^2} - 1} \right)\)

graph of the region

With the help of 3D graphing calculator, obtain the graph of the region, disk and solid as shown below in Figure 1.

In Figure 1, the region is rotated about \(y\)-axis and the solid is obtained by joining the disks along \(y\)-axis.

Therefore, the volume of the solid is \(\frac{{\pi {e^2}}}{2}\left( {{e^2} - 1} \right)\).