Video: Finding the Volume of the Solid Generated by the Revolution of a Region around a Line Parallel to the 𝑦-Axis

Consider the region bounded by the curves 𝑦 = π‘₯Β³, 𝑦 = 0, and π‘₯ = 2. Find the volume of the solid obtained by rotating this region about π‘₯ = 3.

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Video Transcript

Consider the region bounded by the curves 𝑦 equals π‘₯ cubed, 𝑦 equals zero, and π‘₯ equals two. Find the volume of the solid obtained by rotating this region about π‘₯ equals three.

We know that when we rotate a region bounded by a curve 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 𝐴 of 𝑦 with respect to 𝑦, where 𝐴 of 𝑦 is the function that describes the cross-sectional area of the shape. So, it might help to draw a little sketch. We have the curves 𝑦 equals π‘₯ cubed, 𝑦 equals zero, and a vertical line at π‘₯ equals two. And so, that gives us the region we’re looking to rotate. This time though, we’re not rotating about the 𝑦-axis itself but about a line parallel to it. It’s the line with equation π‘₯ equals three.

And so, rotating it about this line, we’d end up with the shape shown. We call this a ring or a washer. And we can say that the cross-sectional area of the ring will be given by the area of the outer, the larger circle minus the area of the inner circle. The area of a circle is πœ‹ times radius squared. And we’re going to rewrite the equation for our curve as π‘₯ in terms of 𝑦, so that’s π‘₯ equals the cube root of 𝑦. And we can say that the radius of the large circle would be the difference between the value of the function π‘₯ equals the cube root of 𝑦 and the function π‘₯ equals three. So, that’s πœ‹ times the cube root of 𝑦 minus three all squared.

Similarly, the radius of our smaller circle will be the difference between three and two. And so, the cross-sectional area or the function for the cross-sectional area is πœ‹ times the cube root of 𝑦 minus three squared minus πœ‹ times three minus two squared. We write the cube root of 𝑦 as 𝑦 to the power of one-third. And we simplify three minus two to be equal to one. The limits of our definite integral are found by considering the equations of the horizontal lines that bound our region. We know the lower of these is 𝑦 is equal to zero. The upper limit is the point of intersection between the curve 𝑦 equals π‘₯ cubed and π‘₯ equals two. So, that’s eight.

We take out a constant factor of πœ‹. And then, we distribute our parentheses. We end up with an integrand of 𝑦 to the power of two-thirds minus six 𝑦 to the power of one-third plus nine minus one. And, of course, nine minus one is simply eight. We then integrate by raising the exponent of each term by one and then dividing by that new value. So, we get πœ‹ times three-fifths of 𝑦 to the power of five over three minus 18 over four times 𝑦 to the power of four over three plus eight 𝑦. We then substitute our limits 𝑦 equals zero and 𝑦 equals eight in. Finally, we simplify. And we find that the volume of the solid obtained by rotating our region about the line π‘₯ equals three is 56πœ‹ over five cubic units.

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