Rather than looking at an example of the washer method with the
$y\text{-axis}$ as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.
The washer method with a different axis of revolution
Find the volume of a solid of revolution formed by revolving the region bounded above by
$f\left(x\right)=4-x$ and below by the
$x\text{-axis}$ over the interval
$\left[0,4\right]$ around the line
$y=\mathrm{-2}.$
The graph of the region and the solid of revolution are shown in the following figure.
We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by
$f\left(x\right)+2,$ which simplifies to
$f\left(x\right)+2=\left(4-x\right)+2=6-x.$
The radius of the inner circle is
$g\left(x\right)=2.$ Therefore, we have
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of
$f\left(x\right)=x+2$ and below by the
$x\text{-axis}$ over the interval
$\left[0,3\right]$ around the line
$y=\mathrm{-1}.$
Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.
Key equations
Disk Method along the
x -axis $V={\displaystyle {\int}_{a}^{b}\pi {\left[f\left(x\right)\right]}^{2}dx}$
Disk Method along the
y -axis $V={\displaystyle {\int}_{c}^{d}\pi {\left[g\left(y\right)\right]}^{2}dy}$
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject.
Well, this is what I guess so.
A conical container of radius 10 ft and height 30 ft is filled with water to a depth of 15 ft. How much work is required to pump all the water out through a hole in the top of the container if the unit weight of the water is 62.4 lb/ft^3?