Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola around the π‘₯-Axis

Find the volume of the solid generated by rotating the region bounded by the curve 𝑦 = βˆ’π‘₯Β² + 2π‘₯ and the π‘₯-axis a complete revolution about the π‘₯-axis.

03:29

Video Transcript

Find the volume of the solid generated by rotating the region bounded by the curve 𝑦 equals negative π‘₯ squared plus two π‘₯ and the π‘₯-axis a complete revolution about the π‘₯-axis.

We recall that the formula we use to find the volume of a solid generated by rotating a region about an π‘₯-axis is the definite integral between π‘Ž and 𝑏 of 𝐴 of π‘₯ with respect to π‘₯, where 𝐴 of π‘₯ is a function which describes the area of the cross section of the volume at a given point. Now, sometimes a nicer formula to use is the definite integral between π‘Ž and 𝑏 of πœ‹ times 𝑦 squared with respect to π‘₯, where π‘₯ equals π‘Ž and π‘₯ equals 𝑏 are the vertical lines that bound our region. And this is the formula we’re going to apply in this question.

Now, to see what’s going on, let’s begin by sketching out the graph of 𝑦 equals negative π‘₯ squared plus two π‘₯. To find the π‘₯-intercept, we’ll set 𝑦 equal to zero and solve for π‘₯. So, that’s negative π‘₯ squared plus two π‘₯ equals zero. Let’s factor an π‘₯ such that π‘₯ times negative π‘₯ plus two equals zero. Now, of course, this statement can only be true if either π‘₯ is equal to zero or negative π‘₯ plus two is equal to zero. And if we solve the second equation by adding π‘₯ to both sides, we find the π‘₯ is equal to two. And so, those are the roots of our equation. They’re the points where the graph intersects the π‘₯-axis.

The equation itself is a quadratic with a negative coefficient of π‘₯ squared. That means it looks like an inverted parabola, as shown. And we’re going to be rotating this region 360 degrees about the π‘₯-axis. Since this region is bounded by the vertical lines π‘₯ equals zero and π‘₯ equals two, we can say the π‘Ž itself must be equal to zero and 𝑏 must be equal to two. So, the volume is the definite integral between zero and two of πœ‹ times 𝑦 squared. Now, 𝑦 is the equation negative π‘₯ squared plus two π‘₯. We can take a constant factor of πœ‹ outside of our integral. And then, the best way to integrate this is simply to distribute our parentheses. When we do, we find our integrand becomes π‘₯ to the fourth power minus four π‘₯ cubed plus four π‘₯ squared. So, let’s perform the integration.

We know that to integrate a polynomial term whose exponent is not equal to negative one, we add one to the exponent and then divide by that new value. This means the integral of π‘₯ to the fourth power is π‘₯ to the fifth power divided by five. When we integrate negative four π‘₯ cubed, we get negative four π‘₯ to the fourth power divided by four. Which simplifies do negative π‘₯ to the fourth power. And then, the integral of four π‘₯ squared is four π‘₯ cubed over three. Now, we need to evaluate this between the limits of zero and two. And so, this becomes πœ‹ times two to the fifth power over five minus two to the fourth power plus four times two cubed over three all minus zero.

This becomes πœ‹ times 32 over five minus 16 plus 32 over three. And then, we’re going to create a common denominator of 15. To do so, we multiply 32 over five by three over three. We write negative 16 as negative 16 over one and then multiply by 15. And we multiply 32 over three by five over five. This is πœ‹ times 96 over 15 minus 240 over 15 plus 160 over 15. Which simplifies fully to 16πœ‹ over 15. And so, we can say that the volume of the solid generated by rotating the region bounded by the curve 𝑦 equals negative π‘₯ squared plus two π‘₯ and the π‘₯-axis 360 degrees about the π‘₯-axis is 16πœ‹ over 15 cubic units.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.