This page covers related rates problems specifically involving volumes where the shape of the volume is described by an equation and is involved in the solution. These types of problems involve cylinders (often called rightcircular cylinders), cones, spheres and troughs (tanks) with a regular geometric shape.
Overview 

When solving these types of problems, you first draw of picture and pick out the type of geometric figure involved. The key here is to write down as many equations as you can think of and then use only the ones that seem to apply. Sometimes, especially with cones, you need to come up with relationships that might not be very intuitive. We will show you some things to watch for. Once you have a figure with all the parts labeled, you can write down the equations involved.
On the main related rates page, we discuss two ways to work these problems based on when you take the derivative, either before combining equations or after. Especially with volume problems, it is often better NOT to combine equations too soon, since the equations can get quite messy very quickly.
What To Do With Constants In Related Rates Problems 

What do you do with constants that are given in the problem? First of all, you never want to just go in and plug in all your constants before you take the derivative.
Safe Answer   Wait and plug in your constants only after you have the derivative. So, you would label all distances with variables, take the derivative with respect to t and then plug in all your given constants. This is what you need to do when you first start learning to work related rates problems. After you have some experience, you can go on to the more experienced technique.
Experienced Answer   Once you learn the basics of related rates problems, you will have a feel for which constants you can plug in right away and which ones you can't. The difference you need to look for is
 if the variable is NOT changing, then you can substitute the constant in before taking the derivative;
 but, if the variable is changing over time, then you must wait until after you take the derivative before you can substitute the constant into the equation.
At this point, it will just confuse you more if we write down a bunch of theory on how to work these problems. You need to actually see one, then work many in order to see patterns. If you haven't already, read the top section (above the resources section) of the main related rates page. Then, come back here, watch a video or two and try your hand at a few problems.
Cylinders 

Problems involving cylinders (also called rightcircular cylinders) and cylindrical tanks are usually pretty straightforward. One of the major variations is the orientation of the tank.
 If the tank is on end, then the volume equation is just the area of the circular crosssection times the height of the tank or the depth of the fluid in the tank, i.e. \(V = \pi r^2 h\).
 The other orientation you may come across is a tank on it's side. In this case, you have to be careful to note if the fluid in the tank is above or below the center line. The equations are different in each case.
Probably, if the problem statement does not say that the tank is on it's side, you can usually assume that it is sitting on end. However, check with your instructor to confirm this.
Problem Statement 

Water is draining out of a cylindrical tank at a rate of 5ft^{3}/min. The diameter of the tank is 8ft and the tank is 10ft tall. How fast is the water level falling when the water is 6ft deep?
Solution 

video by Krista King Math 

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Problem Statement 

Water is being poured into a cylindrical drum at a rate of 0.010m^{3}/sec. If the radius of the drum is 1.2 meters, how fast is the water level rising?
Final Answer 

Problem Statement 

Water is being poured into a cylindrical drum at a rate of 0.010m^{3}/sec. If the radius of the drum is 1.2 meters, how fast is the water level rising?
Solution 

video by Michel vanBiezen 

Final Answer 

\(1/(144\pi)\) m/sec 
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Cones 

Coneshaped tanks are very interesting and you will probably run across more than one problem with a tank shaped like a cone. There are several unique things that you need to understand and watch for when you work related rates problems that involve cones.
1. The formula for the volume of a cone with top radius \(r\) and height \(h\) is \(V = \pi r^2 h/3\).
2. There is a unique relationship that you may not think of when you work these problems. The idea is to take a crosssection of the cone down through the center to the tip of the cone. You end up with a triangle, actually two similar triangles.
 The larger triangle is the size of the tank.
 The smaller triangle is the size of the fluid in the tank.
Check the similar triangles precalculus page for a reminder on how to set up ratios of similar triangles.
3. One trick you need to watch for is whether the cone is point up or point down and what the problem is asking for. If the cone is point down, usually the fluid or whatever is filling the tank is in the shape of a cone, so there is nothing unusual going on IF they are asking for the volume of the fluid. However, if the cone is point up, you need to look at the problem carefully and think about what is going on. As time passes, the shape may or may not stay in the form of an exact cone. Here are a couple of examples.
 If you have something like gravel in a pile, this stays in a cone shape as gravel is being added to the pile. Nothing unusual is going on.
 One trick that I have seen on exams is, if something (like a fluid) is being removed from a coneshape tank when the point is up, there will be some fluid at the bottom and no fluid at the tip (picture on the right). To calculate the volume of the fluid, you need to calculate the volume of the entire cone and subtract the volume of the top cone that is empty. (There are formulas for this volume but what do you do if you don't have the formula on the exam?)
Okay, time for a video. Here is an introduction video for cone problems. The writing is kind of hard to see but I think the discussion will help you get your head around what is going on. I particularly like how he describes the use of the chain rule to calculate the derivative.
video by Khan Academy 

Problem Statement 

Gravel is being dumped from a conveyor belt at a rate of 20ft^{3}/min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10ft high?
Solution 

video by PatrickJMT 

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Problem Statement 

A 10cm tall funnel (point down) is being drained with water at a constant rate of 10cc per second. The mouth of the funnel is 12cm in diameter. How fast is the water level dropping when there are 200cc left in the funnel? [ cc = cubic centimeters ]
Solution 

video by CalculusSuccess 

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Problem Statement 

A funnel is being drained of water at a rate of 0.1ft^{3}/sec. How fast is the water level dropping when there is 4ft of water in the funnel? The full height of the funnel is 9ft and the radius at the top is 3ft.
Final Answer 

Problem Statement 

A funnel is being drained of water at a rate of 0.1ft^{3}/sec. How fast is the water level dropping when there is 4ft of water in the funnel? The full height of the funnel is 9ft and the radius at the top is 3ft.
Solution 

video by Michel vanBiezen 

Final Answer 

\( 9/(160\pi) \approx 0.0179 \) ft/sec 
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Spheres 

Sphere problems show up in two main types.
1. If you have a sphere, like a balloon that is a complete sphere. In this case, you use the equation of the volume of a sphere with radius \(r\) as \(V = 4\pi r^3 / 3\).
2. If you have a spherically shaped tank and it is only partially full of fluid. In this case, you need have to notice if the tank level is above or below the halfway point. The equations are different in each case.
Problem Statement 

Air is being pumped into a spherical balloon at 10cm^{3}/min. Calculate the rate at which the radius of the balloon is increasing when the diameter is 15cm.
Solution 

video by Krista King Math 

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Problem Statement 

A spherical balloon is being filled with air so that the radius is increasing at the rate of 0.5 inches per second. How fast is the volume changing when the radius is 2.5 inches?
Solution 

video by MathTV 

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Problem Statement 

Air is being pumped into a spherical balloon so that its volume is increasing by 400 cm^{3}/sec. How fast is the radius increasing when the radius is 100 cm?
Final Answer 

Problem Statement 

Air is being pumped into a spherical balloon so that its volume is increasing by 400 cm^{3}/sec. How fast is the radius increasing when the radius is 100 cm?
Solution 

video by Krista King Math 

Final Answer 

\(dr/dt=1/(100\pi)\) cm/sec 
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Problem Statement 

A spherical balloon is being filled with air so that its volume is increasing at a rate of 3ft^{3}/min. How fast is the radius increasing when the radius equals 1ft?
Final Answer 

Problem Statement 

A spherical balloon is being filled with air so that its volume is increasing at a rate of 3ft^{3}/min. How fast is the radius increasing when the radius equals 1ft?
Solution 

video by Michel vanBiezen 

Final Answer 

\( 3/(4\pi) \) ft/min 
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Problem Statement 

If we have a sphere whose radius is expanding at a rate of 2cm/sec, how fast is the volume changing when the radius is 10cm?
Final Answer 

Problem Statement 

If we have a sphere whose radius is expanding at a rate of 2cm/sec, how fast is the volume changing when the radius is 10cm?
Solution 

video by Michel vanBiezen 

Final Answer 

\(800\pi\) cm^{3}/sec 
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Problem Statement 

We have a semispherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Hint 

The volume of water in the bowl is given by the equation \(V_w = \pi R h^2(\pi/3) h^3\) where h is the height of the water.
Also, 1L = 1000 cm^{3}.
Problem Statement 

We have a semispherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Final Answer 

Problem Statement 

We have a semispherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Hint 

The volume of water in the bowl is given by the equation \(V_w = \pi R h^2(\pi/3) h^3\) where h is the height of the water.
Also, 1L = 1000 cm^{3}.
Solution 

video by Michel vanBiezen 

Final Answer 

\(5/(6\pi)\) cm/sec 
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Troughs 

These types of problems are troughs (tanks) that have some kind of geometric shape that is the same for the entire length of the trough when you look at the crosssection. The volume equation for these types of tanks is just the area of the crosssection times the length.
Key  It is important to notice that the crosssection of the tank is exactly the same no matter where along the tank you look at the crosssection.
If you have a triangular crosssection, you will end up with similar triangles. See the precalculus similar triangles page for a review on how to set up the ratios for use in related rates problems.
Problem Statement 

We have a 10 meter long trough that we are filling with water at a rate of 0.3 cubic meters per minute. The bottom of the trough is 30 cm wide, the top of the trough is 80 cm wide and it is 50cm tall. What is the rate at which the water level rises when the depth of the water is 40 cm?
Final Answer 

Problem Statement 

We have a 10 meter long trough that we are filling with water at a rate of 0.3 cubic meters per minute. The bottom of the trough is 30 cm wide, the top of the trough is 80 cm wide and it is 50cm tall. What is the rate at which the water level rises when the depth of the water is 40 cm?
Solution 

video by Michel vanBiezen 

Final Answer 

30/7 cm/min 
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For related rates problems involving similar triangles, it may help you to review how to set up the ratios. You can find a discussion of this on the similar triangles precalculus page. 
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