The region under the curve #y=lnx/x^2, 1

Question

The region under the curve #y=lnx/x^2, 1<=x<=2# is rotated about the x axis.  How do you find the volume of the solid of revolution?

Charles Arocha · Accepted Answer

See below.

To find the volume of revolution, we use the idea of a series of discs. We find the volume of each disc and these are summed together. From the diagram, we can see that the rectangle has a height of #ln(x)/x^2# and a width of #Deltax#. This rectangle is rotated around the x axis through an angle #pi# and results in a disc. The volume of this disc will then be:

#V=pi(ln(x)/x^2)^2*Deltaxcolor(white)(88)# ( #V=pir^2h#)

We then sum all the disc in the interval together to find the total volume of the solid. This is the same idea as when we find the area between a curve and the axes.

#:.#

#int_(1)^(2)(ln(x)/x^2)^2dx#

#(ln(x)/x^2)^2=(ln(x)^2)/x^4#

#V=piint_(1)^(2)(ln(x)/x^2)^2dx=-(9(ln(x))^2+6ln(x)+2)/(27x^3)#

#V=pi[-(9(ln(x))^2+6ln(x)+2)/(27x^3)]^(2)-[-(9(ln(x))^2+6ln(x)+2)/(27x^3)]_(1)#

Plugging in the upper and lower bounds:

#V=pi[-(9(ln(2))^2+6ln(2)+2)/(27(2)^3)]^(2)-[-(9(ln(1))^2+6ln(1)+2)/(27(1)^3)]_(1)=0.025542pi#

Volume of revolution:

Charles Arocha · Answer

undefined To find the volume of the solid of revolution formed by rotating the region under the curve $y=\frac{\ln(x)}{x^2}$, where $1\leq x \leq 2$, about the x-axis, we use the formula for the volume of a solid of revolution:

$$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$$

In this case, the function $f(x) = \frac{\ln(x)}{x^2}$ is squared because we are rotating around the x-axis. The limits of integration are $a=1$ and $b=2$. So, the volume can be calculated as follows:

$$V = \pi \int_{1}^{2} \left(\frac{\ln(x)}{x^2}\right)^2 \, dx$$

$$V = \pi \int_{1}^{2} \frac{\ln^2(x)}{x^4} \, dx$$

$$V = \pi \left[-\frac{\ln^2(x)}{3x^3} - \frac{2\ln(x)}{3x^3} + \frac{2}{9x^3}\right]_{1}^{2}$$

$$V = \pi \left[\left(-\frac{\ln^2(2)}{3\cdot 2^3} - \frac{2\ln(2)}{3\cdot 2^3} + \frac{2}{9\cdot 2^3}\right) - \left(-\frac{\ln^2(1)}{3\cdot 1^3} - \frac{2\ln(1)}{3\cdot 1^3} + \frac{2}{9\cdot 1^3}\right)\right]$$

$$V = \pi \left[\left(-\frac{\ln^2(2)}{24} - \frac{\ln(2)}{12} + \frac{1}{36}\right) - \left(0\right)\right]$$

$$V = \pi \left[-\frac{\ln^2(2)}{24} - \frac{\ln(2)}{12} + \frac{1}{36}\right]$$

The region under the curve #y=lnx/x^2, 1&lt;=x&lt;=2# is rotated about the x axis. How do you find the volume of the solid of revolution?

The region under the curve #y=lnx/x^2, 1<=x<=2# is rotated about the x axis. How do you find the volume of the solid of revolution?