Calculus I - More Volume Problems

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Section 6.5 : More Volume Problems

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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.

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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.

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. $A\left( y \right)$ .

Show Step 2

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.

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 $s$ .

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,

$A\left( y \right) = {s^2} = {\left( {2\sqrt {{r^2} - {y^2}} } \right)^2} = 4\left( {{r^2} - {y^2}} \right)$ Show Step 3

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 $y$ .