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 π¦ = 5π₯Β² and π₯Β² + π¦Β² = 2, for π¦ 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 π¦ equals five π₯ squared and π₯ squared plus π¦ squared equals two, for π¦ 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, weβre looking to find the volume of a solid formed by rotating the area between two curves about the π₯-axis. So letβs begin by sketching this area. π¦ 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 π¦ squared equals two is the equation of a circle centered at zero with radius root two.

The region specified is for π¦-values greater than or equal to zero. So, in fact, weβ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 π of a solid generated by revolving the region bounded by two curves, π¦ equals π of π₯ and π¦ equals π of π₯, where π is greater than or equal to π on the interval π, π, about the π₯-axis is given by π equals π times the definite integral with respect to π₯ from π₯ equals π to π₯ equals π of π of π₯ squared minus π 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 theyβre equal. And what weβ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 π, if we rearrange π₯ squared plus π¦ squared equals two to make π¦ the subject, we have π¦ equals the square root of two minus π₯ squared, which is π of π₯. π of π₯ is then five π₯ squared.

Now, the limits of integration π and π 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. Weβ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 π’ equals π₯ squared, we can solve the resulting quadratic for π’. And now making some space for our calculation, using the quadratic formula or otherwise, we have π’ 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 and π. And with our functions π of π₯ and π 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 π¦π§-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 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 π to two decimal places for reasons of space, we have two π 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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