Calculus I - More Volume Problems

Section 6.5 : More Volume Problems

2. Find the volume of the solid whose base is a disk of radius $r$ and whose cross-sections are squares. See figure below to see a sketch of the cross-sections.

$A sketch showing a typical cross-section of the solid. The base of the solid is the disk ${x^2} + {y^2} = {r^2}$, drawn in perspective as an ellipse, and a shaded square stands vertically on the disk with its base running across the disk. The distance from the center of the disk out along the base is labeled $x$ and half the width of the square is labeled $y$.$

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Here are a couple of sketches of the solid from three different angles. For reference the positive $x$-axis and positive $y$-axis are shown.

$Three sketches of the solid from different angles with a typical cross-section shown on each and labeled $A\left( y \right)$. The base of the solid is a disk and the cross-sections taken perpendicular to the y-axis are squares, so the solid looks like a rounded, dome shaped block that is flat on the bottom. For reference the positive x-axis and positive y-axis are shown on each sketch.$

Because the cross-section is perpendicular to the $y$-axis as we move the cross-section along the $y$-axis we’ll change its area and so the cross-sectional area will be a function of $y$, i.e. $A\left( y \right)$.

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While the sketches above are nice to get a feel for what the solid looks like, what we really need is just a sketch of the cross-section. So, here’s a couple of sketches of the cross-sectional area.

$Two sketches of the cross-sectional area. The one on the left is the figure from the problem statement, a square standing on the disk with the left half of the boundary circle colored orange and the right half colored green. The one on the right looks straight down at the base from above and shows the circle ${x^2} + {y^2} = {r^2}$ with the left half labeled $x = - \sqrt {{r^2} - {y^2}} $ in orange and the right half labeled $x = \sqrt {{r^2} - {y^2}} $ in green, and the cross-section itself is the horizontal red line across the circle.$

The sketch on the left is really just the graph given in the problem statement with the only difference that we colored the right/left sides so it will match with the sketch on the right. The sketch on the right looks at the cross-section from directly above and is shown by the red line.

Let’s get a quick sketch of just the cross-section and let’s call the length of the side of each square $s$.

$A sketch of a single cross-section, which is a square with each side labeled $s$. Along the bottom of the square a black dot marks where the y-axis passes through, splitting the base into two pieces each labeled $\frac{s}{2}$, and the two ends of the base are marked with dots and labeled $x = - \sqrt {{r^2} - {y^2}} $ on the left and $x = \sqrt {{r^2} - {y^2}} $ on the right.$

Now, along the bottom we’ve denoted the $y$-axis location in the cross-section with a black dot and the orange and green dots represent where the left and right portions of the circle are at. We can also see that, assuming the cross-section is placed at some $y$, the green dot must be a distance of $\sqrt {{r^2} - {y^2}} $ from the $y$-axis. Likewise, the orange dot must also be a distance of $\sqrt {{r^2} - {y^2}} $ from the $y$-axis (recall we want the distance to be positive here and so we drop the minus sign from the function to get a positive distance).

Now, we know that the area of the square is simply ${s^2}$ and from the discussion above we see that,

\[\frac{s}{2} = \sqrt {{r^2} - {y^2}} \hspace{0.5in} \Rightarrow \hspace{0.5in}s = 2\sqrt {{r^2} - {y^2}} \]

So, a formula for the area of the cross-section in terms of $y$ is,

\[A\left( y \right) = {s^2} = {\left( {2\sqrt {{r^2} - {y^2}} } \right)^2} = 4\left( {{r^2} - {y^2}} \right)\]

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Finally, we need the volume itself. We know that the volume is found by evaluating the following integral.

\[V = \int_{c}^{d}{{A\left( y \right)\,dy}}\]

We already have a formula for $A\left( y \right)$ from Step 2 and from the sketches in Step 1 we can see that the “first” cross-section will occur at $y = - r$ and that the “last” cross-section will occur at $y = r$ and so these are the limits for the integral.

The volume is then,

\[V = \int_{{ - r}}^{r}{{4\left( {{r^2} - {y^2}} \right)\,dy}} = \left. {4\left( {y{r^2} - \frac{1}{3}{y^3}} \right)} \right|_{ - r}^r = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{16}}{3}{r^3}}}\]

Do not get excited about the $r$ integral and area formula. It is just a constant. The only letter that is actually changing is $y$.