So I'm trying to solve this problem but one thing I don't get is, how do I know which method to use just by looking at the graph (disk, washer, or shell)?
Find the volume of the solid generated by revolving the region bounded by the graphs y=x^2- 4x+ 5 and y= 5-x about the line y=-1.
I found the points of intersection between the parabola and the y=5-x, and I get x=3 and x=0. So my a will be 0, and my b will be 3. And on this interval 5-x has greater y values than those of the parabola, so I do: (5-x)-(x^2-4x+5) and I get 3x-x^2. How do I proceed though? I think I get all the steps UP to this point, which is choosing which method to use. Thanks!
discs and washers are the same. Washers are just discs with holes in them. Since x is difficult to express as a function of y, we want to integrate using dx.
Since we're revolving around a horizontal line, using dx means working with discs, since the thickness of each disc will be dx. Each disc will have a hole in it.
v = ∫[0,3] π(R^2-r^2) dx
where R = (5-x)+1 and r = (x^2-4x+5)+1
So,
v = π∫[0,3] ((5-x)+1)^2-((x^2-4x+5)+1)^2) dx
= π∫[0,3] -x^4 + 8x^3 - 27x^2 + 36x
= 162/5 π
If you like pain, you could use shells, integrating on dy. This is awkward, since the parabola has two branches. Its vertex is at (2,1). So, we have to break the integrand up into two parts, one where the height of the shells is the distance between the two sides of the parabola, and the rest where the height is the distance between the curve and the line.
y=5-x means x = 5-y
y=x^2- 4x+ 5 means x = (4±√(16-4(5-y)))/2 = 2±√(y-1)
v = ∫ 2πrh dy in each part, making it
v = 2π∫[1,2] (y+1)((2+√(y-1))-(2-√(y-1))) dy
+ 2π∫[2,5] (y+1)((5-y)-(2-√(y-1))) dy
= 2π(52/15 + 191/15)
= 162/5 π
whew
Wow, thank you so much! But I'm wondering where the + 1 for the R and r, comes into play?
rotated around y = -1, not y=0. So the radius of each disc/shell is y+1, not just y.
Oh, so that value is to be added. Okay, thank you!
To determine which method (disk, washer, or shell) to use when finding the volume of a solid generated by revolving a region about a line, you need to consider the shape of the cross-sections created by the rotation.
First, consider the disk method. This method involves slicing the solid perpendicular to the axis of rotation. If the cross-sections appear as circular disks or cylinders, then the disk method is suitable for finding the volume.
Second, consider the washer method. This method involves slicing the solid parallel to the axis of rotation. If the cross-sections appear as annular rings or washers, then the washer method is appropriate for finding the volume.
Finally, consider the shell method. This method involves slicing the solid parallel to the axis of rotation. If the cross-sections appear as cylindrical shells or rectangles, then the shell method is the most effective for finding the volume.
In your specific problem, you need to find the volume of the solid generated by revolving the region bounded by the graphs y = x^2 - 4x + 5 and y = 5 - x about the line y = -1.
To determine which method to use, you can visualize the cross-sections of the solid. Imagine slicing the solid perpendicular to the y-axis and observe the shape of the resulting cross-sections.
In this case, the resulting cross-sections form cylindrical shells parallel to the y-axis. Therefore, the most appropriate method to use is the shell method.
To continue solving the problem using the shell method, you need to integrate the volume of these cylindrical shells from a to b (where a and b are the points of intersection between the functions).
The volume of each shell is given by the product of its height (the difference between the two functions) and its circumference, which is 2πr.
Using the equation you determined earlier, v(x) = 3x - x^2, representing the volume of each shell at a given x value, you can set up the integral as follows:
V = ∫[a, b] v(x) dx
Substituting the equation for v(x), you have:
V = ∫[a, b] (3x - x^2) dx
Evaluate this integral using the limits of integration a and b to find the volume of the solid generated by revolving the region about the line y = -1.
Note: Make sure to always double-check your integration limits and the setup of the integral to ensure accurate results.