WEBVTT
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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.
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In this example, weβre looking to find the volume of the solid obtained by rotating a region bounded by two curves about the π¦-axis.
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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¦.
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This is sometimes written, alternatively, as the definite integral between π and π of π times π₯ squared dπ¦.
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So, to help us picture whatβs happening, weβre going to begin by sketching the area bounded by the two curves.
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The graph of π₯ equal six minus five π¦ squared looks a little something like this.
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And π₯ equals π¦ to the fourth power looks as shown.
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And so, this is the region weβre going to be rotating about the π¦-axis.
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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.
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Then, when we rotate this region about the π¦-axis, we finally get this rather unusual doughnut shape.
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In fact, we call this a ring.
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The cross-sectional area of the ring will be equal to the area of the outer circle minus the area of the inner circle.
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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.
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Armed with this information, we find that the volume is as shown.
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We take out a constant factor of π and distribute our parentheses.
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And our integrand becomes 36 minus 60π¦ squared plus 25π¦ to the fourth power minus π¦ to the eighth power.
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Then, we integrate term by term.
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The integral of 36 is 36π¦.
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When we integrate negative 60π¦ squared, we get negative 60π¦ cubed divided by three, which simplifies to negative 20π¦ cubed.
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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.
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And finally, the integral negative π¦ to the eighth power is negative π¦ to the ninth power over nine.
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We then substitute one and negative one into this expression.
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And then, we calculate 36 minus 20 plus five minus a ninth minus negative 36 plus 20 minus five plus one-ninth.
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That gives us 376 over nine.
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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.