Question Video: Finding the Volume of a Solid of Revolution Bounded by a Parabola and a Circle | Nagwa Question Video: Finding the Volume of a Solid of Revolution Bounded by a Parabola and a Circle | Nagwa

Question Video: Finding the Volume of a Solid of Revolution Bounded by a Parabola and a Circle Mathematics • Third Year of Secondary School

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Consider the region between the curves 𝑦 = 5π‘₯Β² and π‘₯Β² + 𝑦² = 2, for 𝑦 β©Ύ 0. Find the volume of the solid of revolution obtained by rotating this region about the π‘₯-axis, giving your answer to two decimal places.

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Video Transcript

Consider the region between the curves 𝑦 equals five π‘₯ squared and π‘₯ squared plus 𝑦 squared equals two, for 𝑦 greater than or equal to zero. Find the volume of the solid of revolution obtained by rotating this region about the π‘₯-axis, giving your answer to two decimal places.

In this example, we’re looking to find the volume of a solid formed by rotating the area between two curves about the π‘₯-axis. So let’s begin by sketching this area. 𝑦 equals five π‘₯ squared is a quadratic. So its graph is a parabola with its lowest point at zero. And the function π‘₯ squared plus 𝑦 squared equals two is the equation of a circle centered at zero with radius root two.

The region specified is for 𝑦-values greater than or equal to zero. So, in fact, we’re only concerned with an area above the π‘₯-axis. And the region to be rotated about the π‘₯-axis to form a solid is the shaded region between the two curves as shown in the diagram. The solid obtained by rotating this region might look a bit like a doughnut, but with an indent rather than a hole at its center.

To find the volume of this solid of revolution, we recall that the volume 𝑉 of a solid generated by revolving the region bounded by two curves, 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯, where 𝑓 is greater than or equal to 𝑔 on the interval π‘Ž, 𝑏, about the π‘₯-axis is given by 𝑉 equals πœ‹ times the definite integral with respect to π‘₯ from π‘₯ equals π‘Ž to π‘₯ equals 𝑏 of 𝑓 of π‘₯ squared minus 𝑔 of π‘₯ squared.

In our case, within the specified region, the circle with radius root two is greater than five π‘₯ squared apart from at the points of intersection, where they’re equal. And what we’re actually doing in our integration is summing the areas of the infinitely many cross-sectional vertical disks or washers in between where the two curves meet.

So, now for the function 𝑓, if we rearrange π‘₯ squared plus 𝑦 squared equals two to make 𝑦 the subject, we have 𝑦 equals the square root of two minus π‘₯ squared, which is 𝑓 of π‘₯. 𝑔 of π‘₯ is then five π‘₯ squared.

Now, the limits of integration π‘Ž and 𝑏 are the π‘₯-values where the two curves intersect. And we can find these values in various ways, perhaps using a calculator or graphing software. We’ll do this here by equating our two functions and solving for π‘₯. This means we need to solve 25π‘₯ to the fourth power plus π‘₯ squared minus two equals zero for π‘₯. Substituting 𝑒 equals π‘₯ squared, we can solve the resulting quadratic for 𝑒. And now making some space for our calculation, using the quadratic formula or otherwise, we have 𝑒 equal to negative one plus or minus the square root of 201 all over 50.

Now, since this is equal to π‘₯ squared, we can only use the positive result. And evaluating this, we find π‘₯ squared equals 0.26354 and so on. Next, taking the square root on both sides, we have π‘₯ equal to positive or negative 0.51337 to five decimal places. So these values are our limits of integration π‘Ž and 𝑏. And with our functions 𝑓 of π‘₯ and 𝑔 of π‘₯, we have the integral as shown.

To make things a little easier to manage, we can note that our solid is symmetric about π‘₯ is zero and the 𝑦𝑧-plane. So we can take our lower bound to be zero and double the result. Now, taking the squares of our functions, our integrand is two minus π‘₯ squared minus 25π‘₯ to the fourth power. Integrating each term separately, we then have two πœ‹ times two π‘₯ minus π‘₯ cubed over three minus 25 over five π‘₯ to the fifth power, which is five π‘₯ to the fifth power, all evaluated between π‘₯ is zero and 0.51337.

Substituting our limits of integration in and evaluating, where for π‘₯ is equal to zero each term is zero, and we restrict the value of 𝑏 to two decimal places for reasons of space, we have two πœ‹ times 0.80335 and so on. This evaluates to 5.0476 and so on. And so the volume of the solid of revolution formed by rotating the given region about the π‘₯-axis is 5.05 cubic units to two decimal places.

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