Problem

Find the volume obtained by rotating the region b…

09:08

Problem 62 Hard Difficulty

Find the volume obtained by rotating the region bounded by the curves about the given axis.

$ y = \sin^2 x $ , $ y = 0 $ , $ 0 \le x \le \pi $ ; about the x-axis

Answer

The volume of the solid is $\frac{3 \pi^{2}}{8}$

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

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Video Transcript

this problem is from Chapter seven section to problem number sixty two in the book Calculus Early Transcendental Sze in Tradition by James Door Here we like to find the volume obtained by rotating the region bounded by the curves Why equal sign Square necks y equals zero in between X equals zero and pi and we'd like to rotate this about the X axis. It's over here on the right. You have a rough sketch of the graph y equals sine squared and blue for X between zero and pi and we also have and read the horizontal line Y equals zero. So from these graphs we can see that the area that we're interested in rotating is this area between the blue and the red which weaken to know by our. So after we rotate this thing about the X axis, we have the following another of sketch and here's a cross section and let's do another cross section. So what do you see? That our our cross sections or washers arm sees me disc. So we'd like to find the volume which will be the general of the area of the cross sections. So we have to be more precise here since we obtained the volume of the solid By letting the washers moving the extraction, we'LL have to integrate with respect to X So X is going between zero and pi. Now we need a formula for a in terms of X so is equal to pi r squared because is a disc a circle including the inside of the circle So we have pi r squared area of a circle Now we need to represent our in terms of X. So in Ackley case, we could come back to the diagram that we have and this is one of our cross sections in the center. That's a circle to find the radius. Let's go to the center right here on the x axis and to find the radius. We see it's just a distance from the center, which is why I called zero up into the graph. So we have r is equal to why the length of our here. Why am I not zero? But since why is on the graph sign next clear, we can replace why with sine squared legs. So we have a equals pi r squared, which is pie science squared X Square, and that equals pi. Science of the fourth Power X. So this is the term that will be integrating. And let's pull up that pie outside the integral sign. So picking up where we left off, we have the equals pie in a girl's ear. Opie Science. If we have science to the fourth power, let's go ahead and re write that as science. Where Time Sign square. Then we can use one of the path a grand entities to rewrite us. So let's use the identity. That's deeds. Science for Lex is one minus two, called Cho San to X, all divided by two. So using this formula, we have one minus co. Sign to X over one minus costs and to accept it, too. So let's go ahead and pull this four out of the denominator and simplified the numerator as much as we can. So this true evaluating the numerator, multiplying this out. We have one minus to co sign to X plus co sign squared to X. Now we can help. We can integrate one and co sign of two works to integrate co sign squared of two X. We should use the pathetic and identity, which says co sign Squared X is one plus cause I to x opportune. We have whatever for in a girl's ear in a pie one minus two co signed to X And now using this identity over here on the right, we can write this as one half plus co sign of two times two x so co sign up for X over two. And now we can go out and agree each of these terms. And also note that this one in the one half well combined to three. House What? So what's going on? Integrate this we have power for and a girl of three over two. Three except you and a girl of negative to co sign two works is minus two. Signed to X over, too. An integral co sign up for X is signed for X over four. So we have four times two and we have the end points see, or a pie. So it's Gordon. Come down here to the side of plug. All this in by Over four now was plugging pipe burst three point two, minus sign of two by plus sign of a bi over eight And then when we plug in zero, we see that each of these three terms go to zero. So we subtract nothing, and we could see from the unit circle Sign of two pi zero sign of a pi zero. Some were left with Pyla for times three pi over too. So again, three pies square all over eight and there's our area. That's a volume.

J H.