Find the volume using shell and disk method

[Guest]

Can you please help me ?
Thank you in advance

[Guest]

First, do you know what the "shell" and "disk" methods are?

To find a volume of a figure created by rotating a 2d region about an axis using the "shell method, imagine cutting the region with lines parallel to the axis. Here lines parallel to the axis will be vertical, from the x-axis up to the graph. But that graph "changes" at x= 1 so do this in two different parts. For x= 0 to 1, the lines go from y= 0 up to y= x (the line through (0, 0) and (1, 1) is y= x) so at each x has lenth x- 0= x. further the horizontal distance from x to the axis of rotation, x= -2, is x- (-2)= x+ 2. So, as that rotates around the axis it will sweep out a cylinder with radius x+ 2, so circumference [tex]2\pi(x+ 2)[/tex] and surface area [tex]2\pi(x+2)(y)=2\pi(x^2+ 2x)[/tex]. Taking thickness "dx" the volume of each shell is [tex]2\pi (x^2+ 2x)dx[/tex] and the total volume is the "sum" of all those, or, more correctly, the integral, [tex]2\pi \int_{x= 0}^1 x^2+ 2x dx[/tex]. Now we need to do the same with the other part, from x= 1 to x= 4. Here, each shell would go up from y= 0 to the parabola [tex]x= y^2[/tex] or [tex]y= \sqrt{x}[/tex]. The rest is the same- the radius is x+ 2 so the surface area is [tex]2\pi(x+2)(y)= 2\pi (x^{3/2}+ 2x^{1/2}[/tex] so the total volume is [tex]2\pi \int_1^4 x^{3/2}+ 2x^{1/2} dx[/tex]. The entire volume is [tex]2\pi\left(\int_0^2 x^2+ 2x dx+ \int_1^4 x^{3/2}+ 2x^{1/2}dx\right)[/tex].

[Guest]

Now for the disk method! Here we imagine lines perpendicular to the axis of rotation. They will be parallel to the x-axis so each disk will correspond to a different y value and again we have to do this as two parts.

First take y from 0 to 1. Each "disk" will actually be a "washer", with a hole in the middle but we can just ignore the hole at first then subtract it off. For a given y, from 0 to 1, we can take x going from x= y to x= 4 so of length 4- x. Then entire radius goes from x= -2 to 4 so has length 6 and sweeps out a disk of area [tex]\pi(6)^2= 36\pi[/tex]. But the cente has radius 4- x= 4- y so sweeps out an inner disk of area [tex]\pi (4- y)^2[/tex]. That means each "washer" has area [tex]\pi(36- (4- y)^2)[/tex]. Taking the thickness to be dy, the volume is [tex]\pi\int_0^1 (36- (4- y)^2)dy[/tex].

Now the section from y= 1 to 2. The right end is still x= 4 but now the lower end is [tex]x= y^2[/tex] so the entire radius, of length 6 still, sweeps out disk of area [tex]36\pi[/tex] but the inner radius is [tex]y^2+ 2[/tex] so sweeps out a disk of area [tex]\pi(y^2+2)^2[/tex]. The total volume here is [tex]\pi\int_1^2 (y^2+2)dy[/tex].

Together they give total volume [tex]\pi\int_0^1 (36- (4- Y)^2)dy+ \pi\int_1^2 (y^2+ 2)^2dy[/tex].

With any luck those two those methods will give the same answer!