Question 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 | Nagwa Question 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 | Nagwa

# Question 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 π¦-Axis Mathematics • Third Year of Secondary School

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Find the volume of the solid obtained by rotating the region bounded by the curves π₯ = 6 β 5π¦Β² and π₯ = π¦β΄ about the π¦-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 π¦ squared and π₯ equals π¦ to the fourth power about the π¦-axis.

In this example, weβre looking to find the volume of the solid obtained by rotating a region bounded by two curves about the π¦-axis. And so, we recall that the volume obtained by rotating a region about the π¦-axis, whose cross-sectional area is given by the function π΄ of π¦, is the definite integral between π and π of π΄ of π¦ with respect to π¦. This is sometimes written, alternatively, as the definite integral between π and π of π times π₯ squared dπ¦. So, to help us picture whatβs happening, weβre going to begin by sketching the area bounded by the two curves.

The graph of π₯ equal six minus five π¦ squared looks a little something like this. And π₯ equals π¦ to the fourth power looks as shown. And so, this is the region weβre going to be rotating about the π¦-axis. By either solving the equations π₯ equals π¦ to the fourth power and π₯ equals six minus five π¦ squared simultaneously or using graphing software or a calculator, we find these curves intersect at the points where π¦ equals one and π¦ equals negative one. Then, when we rotate this region about the π¦-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 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 π΄ of π¦ is π times six minus five π¦ squared squared minus π times π¦ 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 and distribute our parentheses. And our integrand becomes 36 minus 60π¦ squared plus 25π¦ to the fourth power minus π¦ to the eighth power. Then, we integrate term by term.

The integral of 36 is 36π¦. When we integrate negative 60π¦ squared, we get negative 60π¦ cubed divided by three, which simplifies to negative 20π¦ cubed. The integral of 25π¦ to the fourth power is 25π¦ to the fifth power divided by five, which simplifies to five π¦ to the fifth power. And finally, the integral negative π¦ to the eighth power is negative π¦ 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 π¦-axis is 376 over nine π cubic units.

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