Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola and the Curve of a Power Function about the 𝑦-Axis

Find the volume of the solid obtained by rotating the region bounded by the curves π‘₯ = 6 βˆ’ 5𝑦² and π‘₯ = 𝑦⁴ about the 𝑦-axis.

02:45

Video Transcript

Find the volume of the solid obtained by rotating the region bounded by the curves π‘₯ equals six minus five 𝑦 squared and π‘₯ equals 𝑦 to the fourth power about the 𝑦-axis.

In this example, we’re looking to find the volume of the solid obtained by rotating a region bounded by two curves about the 𝑦-axis. And so, we recall that the volume obtained by rotating a region about the 𝑦-axis, whose cross-sectional area is given by the function 𝐴 of 𝑦, is the definite integral between 𝑐 and 𝑑 of 𝐴 of 𝑦 with respect to 𝑦. This is sometimes written, alternatively, as the definite integral between 𝑐 and 𝑑 of πœ‹ times π‘₯ squared d𝑦. So, to help us picture what’s happening, we’re going to begin by sketching the area bounded by the two curves.

The graph of π‘₯ equal six minus five 𝑦 squared looks a little something like this. And π‘₯ equals 𝑦 to the fourth power looks as shown. And so, this is the region we’re going to be rotating about the 𝑦-axis. By either solving the equations π‘₯ equals 𝑦 to the fourth power and π‘₯ equals six minus five 𝑦 squared simultaneously or using graphing software or a calculator, we find these curves intersect at the points where 𝑦 equals one and 𝑦 equals negative one. Then, when we rotate this region about the 𝑦-axis, we finally get this rather unusual doughnut shape. In fact, we call this a ring.

The cross-sectional area of the ring will be equal to the area of the outer circle minus the area of the inner circle. Now, since the area of a circle is given by πœ‹ times radius squared and the radius of each circle is given by the value of the function at that point, the area of the cross section 𝐴 of 𝑦 is πœ‹ times six minus five 𝑦 squared squared minus πœ‹ times 𝑦 to the fourth power squared. Armed with this information, we find that the volume is as shown. We take out a constant factor of πœ‹ and distribute our parentheses. And our integrand becomes 36 minus 60𝑦 squared plus 25𝑦 to the fourth power minus 𝑦 to the eighth power. Then, we integrate term by term.

The integral of 36 is 36𝑦. When we integrate negative 60𝑦 squared, we get negative 60𝑦 cubed divided by three, which simplifies to negative 20𝑦 cubed. The integral of 25𝑦 to the fourth power is 25𝑦 to the fifth power divided by five, which simplifies to five 𝑦 to the fifth power. And finally, the integral negative 𝑦 to the eighth power is negative 𝑦 to the ninth power over nine. We then substitute one and negative one into this expression. And then, we calculate 36 minus 20 plus five minus a ninth minus negative 36 plus 20 minus five plus one-ninth. That gives us 376 over nine. And so, we find that the volume of the solid obtained by rotating our region about the 𝑦-axis is 376 over nine πœ‹ cubic units.

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