Why am I getting different answers when I use cylindrical shell method vs disc method?

Question

I have been stuck on this problem for way too long.

I have to solve for the volume of a solid revolved around the y-axis. This is what I have done so far.

Answer 1

Please note: while what I am going to say will probably sound very critical, understand that my comments are intended to get you to change your habits.

The main issue is that even if I were to typeset exactly what you wrote down on paper, it would not help, because your work lacks organization and clarity. You set up integrals and then solve them, but you don't clearly define what they represent. Whoever reviews or grades your work must spend a great deal of extra effort to decipher what those calculations mean, and this is not how you make them happy to evaluate your work.

For complex shapes such as this, you should label using mathematical notation (e.g., $V_1$, $V_2$, etc.) the volumes you are calculating. You should sketch each component separately and label them accordingly. Provide a brief verbal description. Label the axes and the important features of the plot, such as identified coordinates. Label the curves with the corresponding equations that define them. Imagine you are trying to explain yourself to someone who does not know the answer. Doing mathematics is irrelevant if you cannot clearly communicate its meaning. I shall demonstrate these principles as follows. You don't need to make fancy computer graphics: I only did so because it was easier and faster than sketching it on paper, taking a photo, and then uploading it to the computer.

The following figure represents a cross-section of the upper half of the volume $V$ of interest:

The blue surface is the upper portion of the volume of revolution corresponding to the equation $y = \sqrt{x} + 1$ revolved about the $y$-axis for $x \in [0, \alpha]$ and the orange surface is the equatorial portion of the volume of revolution corresponding to the equation $y = (x-1)^2$ revolved about the $y$-axis for $x \in [1, \alpha]$.

Your method of solution seems to involve first computing the volume $V_1$ of the orange solid below (shown only in the $(+,+,+)$ octant):

The red quarter-disk illustrates a representative disk if we were to integrate for this volume with respect to $y$, which is what you appear to have done. Then, you subtract the volume $V_2$ corresponding to a conical-shaped "cap" in the following figure (lavender):

The plum-colored disk is a representative slice, again if we were to integrate for this volume with respect to $y$. However, you did not set up your second integral this way: you integrated with respect to $x$ and used the method of cylindrical shells, as shown in the following figure:

Again, only a quarter of the full cylinder is illustrated.

Either method (disks or cylindrical shells) will work, but the integral expressions will be different and you must pay careful attention to the limits of integration.

Your expression and calculation for $V_1$ is correct:

$$V_1 = 2 \int_{y=0}^\alpha \pi (\sqrt{y} + 1)^2 \, dy = \frac{\pi}{3}\alpha \left(6 + 8 \sqrt{\alpha} + 3\alpha\right). \tag{1}$$ The extra factor of $2$ arises from the need to account for the solid being symmetric vertically.

I could not find a suitable expression in your work for the volume $V_2$ and consequently assumed that the work immediately following $V_1$ was intended to be the calculation, in which case it is incorrect. However, I do not know because I cannot decipher your unlabeled work. The correct volume by cylindrical shells is:

$$V_2 = 2 \int_{x=0}^{\alpha} 2\pi x \left(\color{red}{\alpha} - (\sqrt{x} + 1)\right) \, dx, \tag{2}$$

because the lavender solid is bounded above by the plane at $y = \alpha$, and below by $y = \sqrt{x} + 1$. If you express this volume using disks, it is

$$V_2 = 2 \int_{y=\color{red}{1}}^\alpha \pi \left((y-1)^2\right)^{\color{red}{2}} \, dy, \tag{3}$$ because the solid's lowest point is at $y = 1$, not $y = 0$; and the radial function it is bounded by is $x = (y-1)^2$, but we must square this again since the differential area of the disk is $dV_2 = \pi x^{\color{red}{2}} \, dx$.

Put together, the total volume is $V = V_1 - V_2$, the calculation of which I leave to you as an exercise.

For the sake of completeness, the following figure illustrates a cylindrical shell for $V_1$:

The corresponding integral is

$$V_1 = 2 \left( \pi \alpha + \int_{x=1}^\alpha 2\pi x (\alpha - (x-1)^2) \, dx \right). \tag{4}$$ This is because for $x \in [0,1]$, the volume enclosed is just a cylinder with radius $1$, and height $\alpha$ (for the upper half). Then, for $x \in [1, \alpha]$, the volume enclosed is the usual cylindrical shell with height $\alpha$ minus the parabolic segment $(x-1)^2$.