Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by Given Lines around the 𝑦-Axis | Nagwa Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by Given Lines around the 𝑦-Axis | Nagwa

Question Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by Given Lines around the 𝑦-Axis Mathematics • Third Year of Secondary School

Find the volume of the solid generated by turning, through a complete revolution about the 𝑦-axis, the region bounded by the curve 9𝑥 − 𝑦 = 0 and the lines 𝑥 = 0, 𝑦 = −9, and 𝑦 = 0.

02:07

Video Transcript

Find the volume of the solid generated by turning, through a complete revolution about the 𝑦-axis, the region bounded by the curve nine 𝑥 minus 𝑦 equals zero and the lines 𝑥 equals zero, 𝑦 equals negative nine, and 𝑦 equals zero.

Remember, when we rotate a region bounded by a curve 𝑥 equals some function of 𝑦 and the horizontal lines 𝑦 equals 𝑐 and 𝑦 equals 𝑑 about the 𝑦-axis, we use the formula the definite integral between 𝑐 and 𝑑 of 𝜋 times 𝑥 squared d𝑦. In our case, the horizontal lines we’re interested in are given by 𝑦 equals negative nine and 𝑦 equals zero. So, we’re going to let 𝑐 be equal to negative nine and 𝑑 be equal to zero. The region we’re interested in is bounded by the curve nine 𝑥 minus 𝑦 equals zero and 𝑥 equals zero. Now, we said 𝑥 needs to be some function of 𝑦. So, we make 𝑥 the subject and we find that our equation is 𝑥 equals 𝑦 over nine. That’s this region. And it looks a little something like this when we rotate it about the 𝑦-axis.

Substituting everything we know into our formula for the volume, and we find it’s equal to the definite integral between negative nine and zero of 𝜋 times 𝑦 over nine squared d𝑦. We take out a constant factor of 𝜋 and we distribute our parentheses. And we see, our integrand is now 𝑦 squared over 81. And in fact at this stage, it might also be sensible to take out this common factor of 81. Then, we know that to integrate a polynomial term whose exponent is not equal to negative one, we add one to that exponent and then divide by the new value. So, the integral of 𝑦 squared is 𝑦 cubed over three.

Then, our volume is 𝜋 over 81 times zero cubed over three minus negative nine cubed over three. And, of course, zero cubed over three is zero. We might then choose to write negative nine as negative nine times negative nine squared or negative nine times 81. And this means we can now simplify by dividing through by 81. And then, negative negative nine divided by three is just three. And so, we found that the volume of the solid generated by rotating our region about the 𝑦-axis is three 𝜋 cubic units.

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