Video: Using the Washer Method to Find a Volume of Revolution

The region bounded by the curve defined by 𝑓(π‘₯) = sin π‘₯ and the π‘₯-axis for 0 ≀ π‘₯ ≀ πœ‹ is rotated about the line 𝑦 = βˆ’3. What is the volume of the solid thus generated?

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Video Transcript

The region bounded by the curve defined by 𝑓 of π‘₯ is equal to sin π‘₯ and the π‘₯-axis for zero less than or equal to π‘₯ less than or equal to πœ‹ is rotated about the line 𝑦 equals negative three. What is the volume of the solid thus generated?

To help us work out the volume, let’s begin by sketching our function and the solid generated by the rotation. Our region is bounded by the curve 𝑓 of π‘₯ is equal to sin π‘₯ and the π‘₯-axis, for values of π‘₯ between zero and πœ‹. And this region is to be rotated about the line 𝑦 is negative three. Let’s try and sketch our volume after rotation to see how it looks. If we look at a cross section of our solid sliced vertically, we have what’s called washer. And we’re going to use the washer method to calculate our volume. In our cross section, we have an inner radius and an outer radius. The inner radius is the radius of the hollow center circle. And the outer radius is the radius from the center to the outer circumference of the solid. The center is the line 𝑦 is equal to negative three.

To calculate the volume, we move through the whole solid summing the area of each vertical cross section using integration. The general procedure for finding the volume of a solid of revolution using the washer method is: one, sketch the area and determine the axis of revolution. Our area is in the shape of a washer. And our axis of revolution is 𝑦 is equal to negative three. We then work out the inner radius and the outer radius of our cross section. And from this, we can find the area. The area of our cross section is then πœ‹ times the outer radius squared minus the inner radius squared.

In our case, the inner radius is a constant. That’s the distance from the base of our area to the axis of revolution. The base of our region is on the line 𝑦 equal to zero. And the axis of rotation is 𝑦 is negative three. So our inner radius is three. Our outer radius is a little bit more complicated because this depends on π‘₯. The distance from the axis of rotation 𝑦 is negative three changes as π‘₯ changes. The outer radius is actually three plus sin π‘₯ as our region is bounded by sin π‘₯. So, for example, if π‘₯ is πœ‹ by two, sin π‘₯ is equal to one. And our outer radius, which is three plus sin πœ‹ by two, is three plus one, which is equal to four.

With our inner radius of three and outer radius of three plus sin π‘₯, we can now work out the area of our cross section. This is a function of π‘₯, which is equal to πœ‹ times three plus sin π‘₯ squared minus three squared. And that’s equal to πœ‹ times nine plus six sin π‘₯ plus sin squared π‘₯ minus nine. The nines cancel. So our area is πœ‹ times six sin π‘₯ plus sin squared π‘₯. It’s worth noting here that if the axis of rotation were vertical, we would have area as a function of 𝑦, not π‘₯.

Our third step in calculating the volume is to find the boundaries of our solid. That gives us the limits of integration. These are defined by the boundaries of our function 𝑓 of π‘₯. And in our case, these are zero and πœ‹. We can note again that if the axis of rotation were vertical, these would be 𝑦-values. Our final step in finding the volume is by integrating the area of the cross sections between our limits. In our case, this means the volume is πœ‹ times the integral between zero and πœ‹ of six sin π‘₯ plus sin squared π‘₯ dπ‘₯. Since the integral of the sum of functions is the sum of the integrals, this means the volume is six πœ‹ times the integral from nought to πœ‹ of sin π‘₯ with respect to π‘₯ plus the integral from zero to πœ‹ of sin squared π‘₯ dπ‘₯.

Using standard formula for integrals, this gives us negative six πœ‹ times cos π‘₯ evaluated between zero and πœ‹ plus πœ‹ times π‘₯ over two minus sin two π‘₯ over four evaluated between zero and πœ‹. That is negative six πœ‹ times negative one minus one plus πœ‹ times πœ‹ by two minus zero minus zero minus zero. And this then gives us 12πœ‹ plus πœ‹ squared over two. Since πœ‹ is approximately equal to 3.142, our volume to two decimal places is 42.63.

So the volume of the solid generated by the region bounded by the function sin π‘₯ and the π‘₯-axis for π‘₯ between zero and πœ‹ rotated about the line 𝑦 is negative three is 42.63.

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