00:01
So in this question, i'm going to find the volume of the solid that's generated by revolving the region about the given line.
00:07
I want to find the region in the second quadrant bounded above by the curve y -equal 16 minus x squared, below by the x -axis, and on the right by the y -axis, about the line x -equals 1.
00:24
So let's start by drawing a picture.
00:27
Here i have y equals 16 minus x squared.
00:31
That's a parabola that's opening down.
00:35
Now it's going to have its vertex way up here at 0 comma 16.
00:41
And it's going to have x intercepts of 4 and negative 4.
00:48
So here is my parabola.
00:52
My parabola is y equals 16 minus x squared.
00:56
I'm bounded below by the x -axis.
00:59
And on the right by the y axis, so it's this region here in the second quadrant, and i am rotating this about the vertical line, x equals 1.
01:14
So the easiest thing to do here is the shell method.
01:19
So if i'm doing the shell method, what do we have? we have volume is going to be equal to an integral from a to b.
01:31
Of 2 pi times the radius of each shell, times the height of each shell, all of this dx.
01:45
So first of all, what are my x limits of integration here? my region is extending from x equals negative 4 to 0.
01:56
Now, imagining these shells, i imagine this vertical line at some value of x.
02:04
And here is x equals 1.
02:08
R is the distance from the axis of revolution to that vertical line.
02:14
In this case, that radius is just 1 minus x.
02:21
Then i'm multiplying that by the height of the shell.
02:25
The height of the shell is the distance from y equals 16 minus x squared to y equals 0.
02:32
That distance is 16 minus x squared.
02:36
And so i'm going to have to integrate this.
02:39
So what will i do? well, i'm going to start by pulling that 2 pi out.
02:44
So what do we have? if i pull that 2 pi out, times an integral from negative 4 to 0.
02:51
Now, i'm going to multiply this out.
02:55
1 times 16, 16...