Lesson Video: Surface Area to Volume Ratios Physics

In this video, we will learn how to find the ratio between the surface area and volume of different shapes.

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Video Transcript

In this video, we’re talking about surface area to volume ratios. We’ll study these ratios for common three-dimensional shapes, including cubes and spheres. And we’ll also come to see how these ratios change as shape size changes.

We can get started by reminding ourselves of what the surface area and volume are for cubes and spheres. Given a cube of side length 𝑙, we know that the area of one of the cube’s faces is 𝑙 times 𝑙, or 𝑙 squared. And then, there are six faces over all, four sides and one on top and one on bottom, which tells us that the total surface area of a cube of side length 𝑙, and we’ll call this 𝑆 sub 𝑐, is equal to six times 𝑙 squared. And then, as we think about the volume of this cube, we’ll call that 𝑉 sub 𝑐, we know that that’s equal to the area of one of the cube’s faces, 𝑙 times 𝑙, multiplied by the cube’s depth, which is also 𝑙. So, that the cube’s volume is the side length multiplied by itself three times, 𝑙 cubed.

As a next step, we can compare the cube’s volume to its surface area by taking this ratio. This is equal to 𝑙 cubed divided by six 𝑙 squared, which means that two factors of 𝑙 cancel from numerator and denominator, leaving us with a volume to surface area ratio for a cube of the side length 𝑙 divided by six. In other words, this ratio is proportional to the side length. Note that this is for the ratio of volume to service area, but we could equally consider the inverse, surface area to volume, and still the relation between surface area and volume would remain the same. We just need to keep an eye on which quantity is in the numerator and which is in the denominator.

Knowing this about a cube, we can then consider similar properties for a sphere. Given a sphere of radius π‘Ÿ, the surface area of that sphere, we can call it 𝑆 sub 𝑠, is four times πœ‹ times that radius squared. And then its volume, what we’ll call 𝑉 sub 𝑠, is equal to four-thirds times πœ‹ times π‘Ÿ cubed. And then, once again, we can solve for this ratio, the volume of our shape to its surface area. When we calculate this fraction, we find that not only do two factors of our dimension of interest, in this case the sphere’s radius, cancel out, but so do the factors of four and πœ‹ in numerator and denominator. This leaves us with π‘Ÿ divided by three so that for a sphere, the volume to the surface area of the sphere is proportional to its radius π‘Ÿ.

So far, we’ve been considering shapes whose dimensions of interest, either the side length 𝑙 or the radius π‘Ÿ, are fixed. But it’s interesting to see for a shape whose size changes, how this change can affect the ratio of volume to surface area. For example, say that we have three spheres. The radius of the first is 𝑅, that of the second is twice that, and that of the third is three times 𝑅. We can label these sphere one, two, and three. And recall from earlier that the ratio of each spheres volume to its surface area is equal to its radius divided by three. This means that for sphere one, that ratio is 𝑅 divided by three, while for sphere two, it’s two 𝑅 divided by three, and for sphere three, it’s three 𝑅 divided by three or simply 𝑅.

We see then that as the size of the sphere changes, so does this ratio. It’s increasing as 𝑅 gets larger and larger. And it turns out that the same trend occurs if we consider a cube instead of a sphere. But considering the shape of a sphere, in particular these three we have here, we can see that as the radius increases, sphere volume grows at a faster rate than surface area, while as the radius decreases, sphere surface area actually changes more than volume.

So, as we consider these ratios of volume to surface area, there’s some subtlety involved in how this ratio changes as the size of a given shape changes. Nonetheless, we can always say that in the case of a cube, this ratio is proportional to the cube’s side length, and in the case of a sphere, it’s proportional to the radius. Though we’ve limited our discussion so far to cubes and spheres, the trends that we’re discovering apply in general to three-dimensional shapes. Even if we had some very irregular three-dimensional object, if the overall size of that object changes, say, by being scaled up or scale down, then the shape’s volume will grow at a faster rate than its surface area as that size increases, while when it decreases to a relatively small size, its surface area will change at a faster rate.

Knowing all this about surface area and volume for three-dimensional objects, let’s get some practice now through an example exercise.

The volume 𝑉 of a sphere of radius π‘Ÿ is given by the formula 𝑉 is equal to four-thirds times πœ‹ times π‘Ÿ cubed. The surface area 𝐴 of the same sphere is given by 𝐴 equals four times πœ‹ times π‘Ÿ squared. Which of the following equations correctly gives the ratio of the volume of the sphere to the surface area of the sphere, 𝑉 divided by 𝐴? (A) 𝑉 divided by 𝐴 is equal to one-third π‘Ÿ cubed. (B) 𝑉 divided by 𝐴 is equal to one-third times πœ‹ times π‘Ÿ. (C) 𝑉 divided by 𝐴 is equal to one-fourth times πœ‹ times π‘Ÿ squared. (D) 𝑉 divided by 𝐴 is equal to one-third times π‘Ÿ. (E) 𝑉 divided by 𝐴 is equal to one-third times π‘Ÿ squared.

OKay, so in this exercise, we’re considering a sphere that has a radius π‘Ÿ. We’re told the equations for the volume as well as the surface area of the sphere in terms of its radius. And we want to solve for the ratio of its volume to that surface area. We can do this by writing out the expression for these two terms and then dividing the volume equation by the area one. When we do this, we’re effectively creating one equation out of two fractions. And notice that the fraction on the left is exactly the ratio we want to solve for.

So then, looking on the right-hand side of this equality, let’s see what will cancel out. There’s a factor of four in both numerator and denominator, also a factor of πœ‹. And moreover, we can see that two factors of the radius π‘Ÿ cancel from top and bottom. When we remove all the cancelled factors, we’re left with π‘Ÿ, the radius of the sphere, divided by three. Looking across our five answer options, we see that option (D) expresses the same ratio. The volume of a sphere to its surface area is equal to one-third times its radius.

Let’s look now at a second example exercise.

The graph shows how the volume 𝑉, the surface area 𝐴, and the volume to surface area ratio, 𝑉 divided by 𝐴, vary with the radius of a sphere. Which line represents 𝑉? Which line represents 𝐴? Which line represents 𝑉 divided by 𝐴?

Looking at this graph, we see that both the vertical and the horizontal axes are unlabeled, but that we have these three curves, the red one, the blue one, and the purple one. We’re told that one of these three curves represents how the volume of a sphere varies with its radius. Another one represents how its surface area changes with its radius. And then the third line shows how the ratio of volume to surface area changes with sphere radius. We want to figure out which line is which.

We notice that all three lines start at the origin. And then, for some smaller values of our sphere radius, the purple line is above the red and the blue lines. Eventually though, the blue line crosses the purple line and the red one does as well. And then, as our radius continues to increase, the red line crosses the blue line and begins to grow fastest of all three.

At this point, we can recall the mathematical equations for the volume of a sphere and its surface area. A sphere of radius π‘Ÿ has a volume four-thirds times πœ‹ times π‘Ÿ cubed, which means that its volume is proportional to its radius cubed. And then, the surface area of a sphere is equal to four times πœ‹ times its radius squared, meaning that this quantity is proportional to π‘Ÿ squared. Once we know the formulas for 𝑉 and 𝐴, we can take their ratio and find that it’s equal to π‘Ÿ divided by three, which tells us that this fraction is proportional to π‘Ÿ. We’ll want to focus on these proportionalities as we figure out which line on our graph corresponds to which quantity.

The line that corresponds to 𝑉, the answer to the first part of our question, will be a line that follows a cubic progression. Likewise, the line that represents 𝐴 will follow a quadratic progression. And the one that represents 𝑉 over 𝐴 will be linear. Knowing this, we can look at our graph and note that the red line, even though it starts out below the blue and the purple line, eventually exceeds them both as the sphere radius increases. So then, by the time it leaves this graph we see, the red line has the steepest slope or gradiant. In other words, it’s increasing at the fastest rate. Of the three functional forms we’re looking for, cubic, quadratic, and linear, we know that it’s the cubic form that increases in this manner. So, we’ll say that the line on our graph representing 𝑉 is the red line because as the sphere radius gets larger and larger, this line increases at the greatest rate.

Next, we want to identify which line represents the surface area of the sphere. And as we said, this line will be represented by a quadratic function. Of the lines remaining on our graph, we see that one, the blue line, has a curve to it as a quadratic function would, while the purple line looks to be straight. So then, we’ll say that it’s the blue line that represents the sphere’s surface area as its radius changes. And this leaves us with just one last line, which we expect will represent 𝑉 divided by 𝐴. As we saw, we expect this line to be linear, that is, to have a constant slope or gradient. And indeed, we see that the purple line fits this description. And so, summing up, the red line represents the volume 𝑉, the blue line represents the surface area 𝐴, and it’s the purple line that represents 𝑉 divided by 𝐴.

Let’s look now at one more example exercise.

Consider a cube that initially has side length π‘₯. If π‘₯ increases by a factor of two, by what factor does the volume of the cube change? By what factor does the surface area of the cube change?

Alright, so here’s our cube that we’re told initially has a side length π‘₯. And then, we’re to imagine that π‘₯ increases by a factor of two. This would lead to a cube that looks like this, with side length two π‘₯. When this change happens, we first want to know, by what factor does the volume of the cube change? Now, if we have a cube where 𝑙 is the length of each side, then we can recall that the volume of that cube is equal to 𝑙 cubed. This means that if we call the volume of the cube with side length π‘₯ β€œπ‘‰ sub π‘₯”, then by our equation for the volume of a cube, that volume is equal to π‘₯ cubed, while the volume of the cube with side length two π‘₯, we can call it 𝑉 sub two π‘₯, is equal to two π‘₯ quantity cubed or eight times π‘₯ cubed.

To answer our question, to say by what factor the volume of the cube changes, we’ll want to divide our final cube volume by our initial volume. When we do this, we see that it’s equal to eight times π‘₯ cubed divided by π‘₯ cubed. So, the factors of π‘₯ cubed cancel, telling us that when we double the side length of our cube from π‘₯ to two π‘₯, its volume has increased by a factor of eight. This is the answer to the first part of our question.

In part two, we want to say by what factor does the surface area of the cube change. If we again consider a cube of side length capital 𝐿, then the surface area of that cube is equal to six times 𝐿 squared. This is because each of the cube’s faces has an area of 𝐿 squared and there are six faces in total. So, if we call 𝐴 sub π‘₯ the surface area of our cube when it’s side length is π‘₯, then that’s equal to six times π‘₯ squared. And then, if 𝐴 sub two π‘₯ is the cube’s surface area after its side length has doubled, then that’s equal to six times two π‘₯ quantity squared or six times four times π‘₯ squared, which equals 24π‘₯ squared.

Once more, we’ll want to take the ratio of the final quantity to the initial quantity. When we divide 24π‘₯ squared by six π‘₯ squared, the π‘₯ squareds cancel out. And then, 24 divided by six is four. So, this is our answer to how the surface area of the cube changes with this change in side length. All together then, when we double the side length of a cube, its volume increases by a factor of eight and its surface area grows by a factor of four.

Let’s summarize now what we’ve learned about surface area to volume ratios. In this lesson, after recalling that for a cube with side length 𝑙, the surface area of that cube is six 𝑙 squared, and its volume is 𝑙 cubed, we saw that the ratio of a cube’s volume to its surface area is its side length divided by six, in other words, that this ratio is proportional to the side length. Related to this, we saw that for a sphere of radius π‘Ÿ, its surface area being four πœ‹π‘Ÿ squared and its volume being four-thirds πœ‹π‘Ÿ cubed means that the ratio of a sphere’s volume to its surface area is equal to its radius divided by three, that is, that this fraction is proportional to the radius.

We saw that as the size of a shape changes, say, by increasing in radius, the ratio of its volume to surface area changes as well. As the shape’s characteristic dimension increases, so does this ratio, while as the shape size decreases, the ratio follows suit. And lastly, we saw that these trends describe not just cubes and spheres, but three-dimensional shapes in general.

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