Video: Finding the Volume Obtained by Rotating a Region About the 𝑦-Axis

What is the volume of the solid formed when the region bounded by the 𝑦-axis, 𝑦 = 𝑒^(βˆ’π‘₯), and 𝑦 = 3 is rotated around the 𝑦-axis?

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Video Transcript

What is the volume of the solid formed when the region bounded by the 𝑦-axis, 𝑦 is equal to 𝑒 to the power of negative π‘₯, and 𝑦 is equal to three is rotated around the 𝑦-axis?

Let’s begin by sketching a diagram of our shape. We know that the curve 𝑦 equals 𝑒 to the power of negative π‘₯ is a reflection of the curve 𝑦 equals 𝑒 to the power of π‘₯ in the 𝑦-axis. It intersects the 𝑦-axis at the point with coordinates zero, one. We also have the line 𝑦 equals three, and it’s bounded by the 𝑦-axis as well. So, the region shaded is the region we’re going to rotate about the 𝑦-axis. When we rotate this region about the 𝑦-axis by 360 degrees, we get a shape that looks a little bit like the one shown. This is called a solid of revolution because it’s a shape that’s obtained by revolving a region about a line.

In general, we calculate the volume of a solid of revolution formed by rotating a region about the 𝑦-axis using the formula 𝑉 is equal to the definite integral between 𝑐 and 𝑑 of 𝐴 of 𝑦 with respect to 𝑦, where 𝐴 of 𝑦 is the cross-sectional area of the shape perpendicular to the 𝑦-axis. Now, since the cross-sectional shape is a circle and we know that the area of the circle is πœ‹π‘Ÿ squared, we can write this as the integral of πœ‹ times π‘₯ squared with respect to 𝑦. And that’s because the radius of the circle will be given by the value of π‘₯ at a given point for 𝑦. Now in this case, our region is bounded by the horizontal lines 𝑦 equals three and, actually, 𝑦 equals one. So, 𝑑 is equal to three and 𝑐 is equal to one.

And so, we’re going to integrate between one and three πœ‹ times π‘₯ squared with respect to 𝑦. Now, the problem we have at the moment is that our curve is 𝑦 equals 𝑒 to the power of negative π‘₯. It’s an equation for 𝑦 in terms of π‘₯, so let’s rearrange. We can take the natural log of both sides of this equation. Now, the natural log of 𝑒 to the power of negative π‘₯ is just 𝑒 to the power of negative π‘₯. So, we find the natural log of 𝑦 is equal to negative π‘₯, which means π‘₯ is equal to negative the natural log of 𝑦. We then replace π‘₯ in our integrand with this expression, with negative the natural log of 𝑦. And we see we’re looking to integrate between one and three πœ‹ times the negative natural log of 𝑦 all squared with respect to 𝑦. But of course, when we square a negative, we get a positive. So, it’s actually πœ‹ times the natural log of 𝑦 squared with respect to 𝑦.

Now, we can actually integrate πœ‹ times the natural log of 𝑦 squared with respect to 𝑦 by using integration by parts more than once. But this question allows the use of a calculator, and we have functions on our calculator that will work this out for us. Typing the definite integral between one and three of πœ‹ times the natural log of 𝑦 squared with respect to 𝑦. Or, alternatively, depending on the functionality of the calculator, we might type the definite integral between one and three of πœ‹ times the natural log of π‘₯ squared with respect to π‘₯. Either way, we get 3.23324 and so on, which, correct to three decimal places, gives us a volume of 3.233 cubic units.

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