How do you find the volume of a solid that is enclosed by #y=1/x#, x=1, x=3, y=0 revolved about the y axis?

Question

Emma Aldridge · Accepted Answer

#4pi# The shape that we are integrating should look something like a big top for a circus. We may have to split up the top and the bottom halves at # y= 1/3#. We can use an integration over #y# here. #V = int_0^1 A(y)dy # To find the area of a slice we look at points below #y=1/3# and points above. Below The outer radius is constant at 3. There is an inner radius cut out from the #x=1# boundary, i.e. the area is the annulus between #r = 1# and #r = 3#. We know this area: #A = pi * 3^2 - pi * 1^2 = 8pi # Above The bound is a little more complicated above #y = 1/3#. In this region, the inner radius is still a constant (hence a #-pi# term we will add) but the outer radius goes like #1/y#, i.e. #A(y) = pi x^2 - pi = pi/y^2 - pi # We can integrate this whole thing piecewise: #V = int_0^1 A(y) dy = int_0^(1/3) A(y) dy + int_(1/3)^1 A(y)dy # #int_0^(1/3) A(y) dy = 1/3 * 8pi = (8pi)/3 # #int_(1/3)^1 A(y) dy = -piy^-1 - pi y = -pi(1 - 3) - pi(1 - 1/3) = 2pi - 2/3 pi = (4pi)/3 # Therefore, the total volume is #V = (8pi)/3 + (4pi)/3 = 4pi #

Ezekiel Alexander · Answer

undefined To find the volume of the solid formed by revolving the region bounded by $y = \frac{1}{x}$, $x = 1$, $x = 3$, and $y = 0$ about the y-axis, you can use the method of cylindrical shells. The formula for the volume of the solid using cylindrical shells is:

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

Where $f(x)$ represents the function defining the curve, and $a$ and $b$ are the bounds of integration.

In this case, $f(x) = \frac{1}{x}$, and the bounds of integration are $x = 1$ and $x = 3$. Thus, the volume can be calculated as follows:

$$V = 2\pi \int_{1}^{3} x \cdot \frac{1}{x} \, dx$$

After simplifying the integral, you'll find that:

$$V = 2\pi \int_{1}^{3} dx$$

$$V = 2\pi \cdot (3 - 1)$$

$$V = 4\pi$$

So, the volume of the solid formed by revolving the given region about the y-axis is $4\pi$ cubic units.