00:01
We are going to find the volume obtained by rotating the region bounded by the curves about the given axis.
00:11
The curves are y -sign of x, y -cosine of x, for x between 0 and pi -fourth, and the rotation would be about y -equal 1.
00:23
So here we have a sketch of the geometry of the problem.
00:29
We have the red curve, this y -sign of x.
00:33
Is one here.
00:39
The blue curve here is y -cosin of x.
00:45
They are equal at pi -fourth, and the region over which we can see there the rotation is between x equals zero and x equal pi -fourth here.
01:06
So in that interval, the area between, or the region between the two curves is shaded in blue here.
01:15
So to calculate the volume of the solid that is generated when we rotate this area, this region, about this axis here, which is horizontal line y equals 1, that is a line, horizontal line passing through the value y equal 1.
01:38
You can see that this line is tangent to the cosine curve at 0 and what we do actually is is to subtract two volumes to obtain the volume we want.
01:57
That is, you have an outer radius, let's say this one.
02:11
So when we rotate up to this value by fault here, around this line, y -equal 1, we get a solid that includes this blank part here, taking this radius that i draw here.
02:33
But we want that region without the blank part.
02:41
So what we do is subtract another volume that is generated with this inner radius here.
02:54
Of course, it must be a vertical line.
02:59
Let's say this way.
03:01
Okay.
03:03
And the solid, this inner radius here will generate a solid that corresponds to the blank, area here and when we subtract that to the previous solid we get the solid generated but this region that is shaded in blue here and what are these radius the first one this radius here corresponds to one minus let's say radius one is one minus the value of the red line at a specific values of x, and that's the sine function.
03:53
And the second radius here, r2, will be 1 minus the value of the blue line at x, that is, cosine, of x.
04:10
So this is the outer radius, this is the inner radius.
04:26
And the subtraction is outer radius minus inner radius.
04:32
Of course i'm talking about subtraction of the areas of the circles.
04:39
With the integration over 0 pi -fault, we get the volume.
04:43
So we can say then that the volume of the solid that is generated by rotating this shaded area here around or about the horizontal line y -e -e -e -e -e -go -1, would be the integral from 0 to pi -fourth of pi times the square of the outer radius, that is r1 square or 1 minus.
05:20
Let me put the formula with the radius, and then after that i put the corresponding formula.
05:30
So it's external radius square minus internal radius square.
05:37
And the x is inside these two values.
05:40
So we see that what we are doing in fact is subtracting the two volumes.
05:46
If we separate these two integrals, we have two volumes and the subtraction of those two volumes give us the volume we want.
05:55
Okay, so that's it...