Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by the Curves of Sine and Cosine Functions about a Line Parallel to the 𝑥-Axis | Nagwa Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by the Curves of Sine and Cosine Functions about a Line Parallel to the 𝑥-Axis | Nagwa

# Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by the Curves of Sine and Cosine Functions about a Line Parallel to the π₯-Axis Mathematics

Find the volume of the solid obtained by rotating the region bounded by the curves π¦ = sin π₯, π¦ = cos π₯, π₯ = π/6, and π₯ = π/4 about π¦ = β1. Give your answer to two decimal places.

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### Video Transcript

Find the volume of the solid obtained by rotating the region bounded by the curves π¦ is equal to sin π₯, π¦ is equal to cos π₯, π₯ is equal to π by six, and π₯ is equal to π by four about π¦ is equal to negative one. Give your answer to two decimal places.

Weβre given that the functions π¦ is equal to sin π₯ and π¦ is equal to cos π₯ form the boundaries of a region and where the region is also bounded by the lines π₯ is equal to π by six and π₯ is equal to π divided by four. And weβre asked to find the volume of the solid obtained by rotating the bounded region about the line π¦ is equal to negative one.

It can be helpful if we first try and sketch our solid. In fact, the solid is a ring, where the cross sections of the band are wedge shaped. Now, to find the volume of this solid, we sum using integration the areas of all its vertical cross sections. The cross sections are actually circular washers. Thatβs where the outer edge is bounded by the cos of π₯ and the inner edge by sin π₯. And to find the area of these cross sections, since the area of a circle is π times the radius squared, letting π sub O, thatβs π outer, be the radius of the outer circle and π sub I, thatβs π inner, be the radius of the inner circle, then the area of a cross-sectional disk π΄ subscript D is equal to π times π outer squared minus π times π inner squared. And of course, we can take the common factor of π outside some parentheses.

Now, recalling that the center of our rotation is at π¦ is equal to negative one, then the outer radius, π subscript O, is the distance from the center of rotation to the line π¦ is equal to zero, thatβs the π₯-axis, which is equal to one, plus the cos of π₯ for the given value of π₯. And so the outer radius is one plus the cos of π₯. And similarly, the inner radius, π I, is the distance from the center of rotation to the line π¦ is equal to zero, so thatβs one, plus the sin of π₯. And this means that the inner radius, π I, is one plus sin π₯. And so the area of a cross-sectional disk is given by π multiplied by one plus the cos of π₯ squared minus one plus the sin of π₯ squared.

Expanding the inner parentheses, we see that the ones disappear. And we have π multiplied by cos squared π₯ minus sin squared π₯ plus two cos π₯ minus two sin π₯. And recalling from our trigonometric identities that the cos squared of π₯ minus the sin squared of π₯ is equal to the cos of two π₯, we have the area of a cross-sectional disk is then π multiplied by the cos of two π₯ plus two cos π₯ minus two sin π₯.

Now, making some room, and now to find the volume of our solid, we integrate the area of these disks with respect to π₯ between our boundaries for π₯. Thatβs where the lower bound is π by six and the upper bound is π by four. So weβre effectively taking the sum of the areas of infinitely many cross-sectional disks bounded by the two functions cos π₯ and sin π₯ and between π by six and π by four. By the additive property of integrals, we can split our integral into three. And we can take the constant factors of π and two π out front. And we have three definite integrals that we know how to evaluate.

In our first integral, weβll use the fact that if π’ is equal to ππ₯, then the integral of the function π of π’ with respect to π₯ is given by one over π times the integral of π of π’ with respect to π’. In our case, π of π’ is equal to the cos of two π₯, where π is equal to two. For our first term then, we have π multiplied by one over two times the sin of two π₯ evaluated between π by four and π over six. We use the fact that the definite integral of the cos of π₯ between π and π with respect to π₯ is the sin of π₯ evaluated between π and π. Our second integral is then two π multiplied by the sin of π₯ evaluated between π over four and π over six. And finally, for our third term, we use the fact that the integral of the sin of π₯ is the negative cos of π₯. And so our third integral is negative two π multiplied by the negative cos of π₯ evaluated between π over four and π over six.

Now, since the limits of integration were the same for each term, we can collect everything together and take a factor of π by two outside some parentheses so that we have the volume is π by two multiplied by the sin of two π₯ plus four times the sin of π₯ plus four times the cos of π₯ all evaluated between π by four and π by six. And now making some space and rewriting, we can substitute in our limits and make some simplifications so that the first term in the first set of parentheses is the sin of π over two and the first term in the second set of parentheses is sin of π over three.

And now evaluating our sines and cosines, the sin of π over two is equal to one. The sin of π by four is equal to root two over two, as is cos of π over four. The sin of π over three is root three over two, and the sin of π over six is one-half. And the cos of π over six is root three over two. So now we see that some of our twos cancel. And we have π over two multiplied by one plus four times the square root of two minus five root three over two plus two. And once again making some space, we have that the volume is π over two multiplied by negative one plus four times the square root of two minus five times root three over two.

Evaluating inside our parentheses to five decimal places, we have π over two multiplied by 0.32673. This evaluates to 0.51 to two decimal places. And so we find that the volume of the solid obtained by rotating the region bounded by the curves π¦ is equal to the sin of π₯, π¦ is the cos of π₯, π₯ is π over six, and π₯ is π over four about the line π¦ is negative one is 0.51 units cubed to two decimal places.