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Q4E

Expert-verifiedFound in: Page 378

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**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**

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

**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**

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

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

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

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

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

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