Find the volume obtained by rotating the region bounded by...

Key Concepts

Trigonometric Integration Techniques

When the function involved is trigonometric, such as sin²x, specific integration techniques like power-reduction formulas are often applied to simplify the integrand. These techniques transform the integrals into forms that are easier to evaluate over the given intervals.

Definite Integration

Definite integration is used to sum infinitesimal contributions to a total quantity—in this case, volume—over a specified interval. It involves setting the correct limits of integration, which are often determined by the boundaries of the region being rotated.

Disk Method

The disk method is a technique used to compute the volume of a solid of revolution when the slices perpendicular to the axis of rotation form disks or circular washers. It involves setting up an integral where each disk's volume is represented by the area of the disk (? times the square of the radius) multiplied by an infinitesimal thickness.

Volume of Revolution

This concept involves creating a three-dimensional solid by rotating a two-dimensional region around an axis. The resulting structure's volume can be computed by integrating small volume elements that make up the solid, which are generated by the rotation of the region's slices.

Transcript

00:02 We will find the volume obtained by rotating the region bounded by the curves y equals sine square of x and y equals 0 for x between 0 and pi about the x axis.

00:19 And here we have a sketch of the function sine square of x which is the red line over here y equal sine square of x and that's a positive function on the interval 0, 0 ,000, pi you have zero here and the pi's over here at zero and at pi the function is zero and at pi half over here is one so for any value of x between zero and pi we take this radius here that is the value of the function at x we calculate the area of these circle with this radius that will give us let's say the part of the solid that is generated but this line.

01:29 If we integrate over 0 pi, we get the whole volume of the solid.

01:34 So let's say then that the volume of the solid generated by rotating the curb sine square of x about the x -axis is that is retaining this area here.

01:59 Is the interval from 0 to pi of pi times sine square of x which is the value of the function square.

02:12 That is the volume is given by the interval from 0 to pi of pi times which can get out of the interval so we can put it out already interval from 0 to pi of sine to the 4th of x.

02:32 So we need to calculate this integral here and after that we multiply by pi we get the volume.

02:40 So let's calculate that internal.

02:55 Okay, so to do that, let's first write the interval.

03:03 Let's use directly the formula of the sine square of an angle.

03:08 For that we start with the formula of cosine of the sum of two angles.

03:15 We know cosine of a plus b is cosine of a, times cosine of b minus sine of a times sine of b and so if we put the same angle here we get cosine of 2a which is cosine of a plus a is equal to put in b equal a here we get cosine of a times cosine of a minus sign of b sign of a times sine of a.

04:09 And then this is, or this implies that cosine of 2a is cosine square of a minus sine square of a.

04:24 And here we replace cosine square of a by one minus sine square of a, which is or which comes from the fundamental identity, trigonometric identity, sine square of an angle, plus cosine square of the same angle equal 1.

04:41 So this cosine square of a is 1 minus sine square of a and then that minus sine square of a.

04:52 We get then that cosine of 2a is 1 minus 2a and from here we can solve for sine square of a.

05:09 So sine square of a is is 1 minus cosine of 2a over over 2.

05:27 And this is the expression we will use to calculate the interval.

05:32 But not seriously because we got to say that we need the square of that.

05:44 That is the interval from 0 to pi of sine to the 4th, which is the interval we want to calculate.

05:55 Is the integral from 0 to pi of sine square of x, sine square of x, square of x, square.

06:14 And then sine square of x can be replaced by an expression like this, and this will become the integral from 0 to pi of 1 minus cosine of 2x over 2 squared.

06:36 Squared because sine square of x putting a equal x we get 1 minus consens of 2 8 2 x over 2 and that is square and now we can develop the square and 1 over 2 square is 1 4th that get out of the integral so is 1 4th times the interval from 0 to pi of the square of 1 minus consign as 2 x and that is 1 minus 2 cosine of 2x okay plus cosine square of 2x now we can separate this once 1 fourth times the integral from 0 to pi of differential of x minus the interval from 0 to pi cosine of 2x, cosine of 2x, plus the intro from 0 to pi of cosine square of 2x.

08:00 Good.

08:01 We have that.

08:05 Let me see if i can now.

08:08 And do this maybe.

08:14 Get this back here and put this back here.

08:22 And then this bigger.

08:25 Okay, we have that.

08:31 And now the first two integral are easy to calculate.

08:35 So we get, this is the length of the interval of integration, pi, minus.

08:41 Here, the primitive is, or a primitive is sine of 2x.

08:47 Because the derivative of sine of 2x is cosine of 2x times 2.

08:51 It's just what we have inside into.