# How to find the work required to pump the water out of the tank

In this post, I am going to be showing you how to find the work required to pump the water out of the spout.  Work is a common topic in calculus 2, and there are a lot of different applications for it.  But finding the work required to pump the water out of a tank is one of the most common applications.

It is also important to keep in mind that the process is slightly different depending on whether your tank is measured in feet or inches versus meters.  So, I will show you how both work out.  But keep in mind that the process is extremely similar except for one main difference, which I will talk more about more soon, and the units of course.

## Find the work required to pump the water out of the spout – Meters

If you are given some sort of a tank like in this case, a spherical tank that we can see has a radius of 3 meters.  And then there is the spout sticking out the top of it, which is 1 meter long.

This is going to use one of the formulas mentioned on my calculus 2 study guide.  You can click here to check that out, but you can go download that right away.  It is available for instant download; it will be a PDF that you can print or save.  It is only a couple bucks, so it is very affordable, but this is one of the formulas on there.

### The General Strategy

When it comes to figuring out the work required to pump water out of a tank, really what that comes down to is setting up an integral which represents the work required to do this.  But when you are setting that integral up, you have to go about it step-by-step and work your way through the different steps.  We are not going to go straight to the formula for work quite yet.  What we want to do first is imagine, as water is being pumped out of this tank, what is going to be happening is, you can imagine a bunch of little disks.  You can see one such disk in the image below.

Each little disk is basically an infinitely thin cylinder, which represents all the water that sits in that layer of our tank.  So, you can kind of imagine, instead of thinking of this as a sphere, think of it as a bunch of really thin cylinders stacked on top of each other that make the shape of a sphere.  And the reason why we want to think of it like this is each disk of water in this tank, is going to require a certain amount of work since all the water in that layer must be pumped up the same distance.  All the water on this layer is going to be the same distance from the top of the spout where we are trying to get that water to.  So basically, what we are going to be doing, is figuring out an equation which represents the amount of work required to pump a specific disk out of this tank.  And then we are going to sum up all those works that it takes to get each disk, and that be the work that it takes to get all the disks of water out of the tank.

### Draw a Sketch and Label Your Variables

To do this, what we want to do, is come up with a new variable, which represents the distance that any given layer of water has to be pumped out.  That distance is going to be the distance from whatever layer of water we are looking at up to the top of the spout, which is the distance labeled $x_i^*$.

And you can imagine, these different disks as we go throughout this whole thing, are all going to have a different distance that they need to be lifted.  The disk down at the bottom of the tank is going to have to be pumped further than the disk up at the top of the tank.  Once we have this variable that represents the distance that the ith layer has to be pumped up, what you want to do is figure out an equation for the volume of that layer.

### Volume of the ith Layer

We want to come up with an equation for the ith layer of the liquid.  This is going to be based on $x_i^*$ because, you can see as you go throughout this sphere, the radius of the individual disk changes.  The distance from the center of the disk to the edge of the disk is much shorter at the top and the bottom of this sphere, and much longer in the middle of the sphere.  In the middle, we know it is 3 meters, but closer to the top or bottom of the tank it is going to be shorter. The volume of each disk is going to be different depending on how far down into the tank you are.

Because the volume of each individual disk is just going to come from the volume of a cylinder equation, we can start with that as a template and adjust from there.

$$V=\pi r^2 h$$

The radius is going to depend on how far down into the tank we are.  So, what we need to do is come up with an equation which depends on $x_i^*$, and represents the radius of that specific disk sitting at the depth of $x_i^*$.  To do that we can graph this circular shape and spout on an x-y-axis.  We will say that the top of the spout is at the origin and will graph the depth from the top of the spout ( $x_i^*$) on the x-axis, and the radius of the tank at that depth (r) on the y-axis.

What we want to do is figure out an equation for that circle.  Generally, the equation for a circle is going to be $(y-y_0)^2+(x-x_0)^2=r^2$.  Since the center of this circle is at (4, 0) and the radius of the tank is 3 m, we can plug in these values to see that the equation of this circle is

$$(y-0)^2+(x-4)^2=3^2$$

Then we can simplify this equation and solve for y, leaving us with

$$y=\pm \sqrt{9-(x-4)^2}$$

So if we want a formula for the radius of our disk we can just take the positive square root piece, since that represents the top half of the circle.  Therefore, the radius is going to be $y=\sqrt{9-(x-4)^2}$.  But we’ll want to change the x to $x_i^*$. So the volume of our ith disk is just going to be

$$V_i = \pi \bigg( \sqrt{9-(x_i^*-4)^2} \bigg) ^2 \ h$$

Now we need to figure out the height. Well, the height of each of these disks will simply be $\Delta x$.  All that means is the distance that you go from one disk, or one layer, of your water to the next.  If we make this change, that tells us

$$V_i = \pi \bigg( \sqrt{9-(x_i^*-4)^2} \bigg) ^2 \ \Delta x$$

And we can simplify this a bit.

$$V_i = \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

Once we have the volume of the ith disk, what we needed to do is figure out the mass of that disk.

### Mass of the ith Layer

Once we have found an equation for the volume of the ith layer of water in the tank, finding the mass of the ith layer is fairly straight forward. The reason for this is that the mass of the layer will just be the volume of the layer times the mass of water. Since we know the mass of water to be $1000 \frac{kg}{m^3}$, this tells us that

$$m_i = 1000 * V_i$$

$$m_i = 1000 \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

### Force Acting on the ith Layer

Again, finding the force acting on the ith layer is pretty simple once we know the mass of the ith layer. This is because the force is mass times gravity. Or in other words

$$F_i = g * m_i$$

Since we know the gravitational constant for the strength of gravity on Earth is $9.8 \frac{m}{s^2}$, we just have to multiply the mass of the ith layer by 9.8 to get the force acting on the ith layer.

$$F_i = 9800 \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

### Work Required to Pump the ith Layer Out of the Spout

Now that we know the force acting on the ith layer, we can use this to find the work required to pump the ith layer of water up and out of the spout. This is where the formula on my calculus 2 study guide comes in. That formula says that work is force times distance.

Since we already know the force of the ith layer, the only thing left to figure out is the distance that the ith layer needs to be pumped. Well, if we look back up to the drawing where we named our variables, you will see that $x_i^*$ represents the distance between the ith layer of water and the top of the spout. So $x_i^*$ is the distance that the ith layer needs to be pumped. So if we know that

$$Work \ = \ Force \ * \ Distance$$

And that $x_i^*$ is the distance that the ith layer needs to be pumped, then

$$W_i = F_i \ * \ x_i^*$$

$$W_i = 9800 x_i^* \pi \Big( 9-(x_i^*-4)^2 \Big) \ \Delta x$$

### Total Work

At this point we only know the work required to lift the ith layer, not the work that is required to lift all the water out.  What we need to do now is use this to set up an integral which would give us the total amount of work to lift all the i layers, which is all the water.  To do that we can set up an integral. And we are going to integrate this work that it takes to lift the ith layer.  Basically, what the integral is doing is summing up all the amounts of work that it takes to lift each individual layer. It is adding all those up together to give us one total amount of work that it takes to pump all the water out of the spout.

When you put it in terms of an integral, this $\Delta x$ changes to a dx.  We would also need to change the $x_i^*$ to x.  This change of variable just changes us from being in the context of a sum to the context of an integral, which is just a special type of an infinite sum itself.

$$\int 9800 x \pi \Big( 9-(x-4)^2 \Big) \ dx$$

### But what about the bounds?

Now what we need to figure out is the bounds of our integral.  The bounds are going to represent the range of these x values, that we want to integrate over.  To do this we want to think about where the x came from. Again, this was the distance that we must pump our water.  What we need to figure out is: what is the range that we have to pump each layer of water?  What are the total range of distances?

Well, we can see looking back up to our drawing, that the top layer of water is already at the bottom of the spout.  It just has to be pumped up 1 meter to get to the top of the spout.  So, the shortest distance we are going to have to pump is 1 meter. That tells us that the lower bound of our integral is going to be 1.

The very, very bottom bit of water down at the bottom of our tank is going to have to be pumped all the way up through the tank and then all the way up through the spout.  If the radius of our sphere is 3 meters, it is going to have to go 3 meters to get up to the center of our sphere, another 3 meters to get up to the top of the sphere, and then another 1 meter to get to the top of the spout.  So, that is going to be $3 + 3 + 1 = 7$ total meters.  That tells us that the upper bound is going to be 7.

$$\int_1^7 9800 x \pi \Big( 9-(x-4)^2 \Big) \ dx$$

So now we can integrate this and that would give us the total work that it would take to pump all the water out of the spout.

$$9800 \pi \int_1^7 x \Big( 9-(x-4)^2 \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( 9-(x-4)(x-4) \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( 9-(x^2-8x+16) \Big) \ dx$$

$$9800 \pi \int_1^7 x \Big( -x^2+8x-7) \Big) \ dx$$

$$9800 \pi \int_1^7 -x^3+8x^2-7x \ dx$$

This can be done using the power rule, then evaluating over our bounds of 1 to 7, we get $1,411,200 \pi$. So, we know that it would take $1,411,200 \pi$ Joules worth of work to pump all of the water out of the spherical tank.

So, like I said, this is one of the formulas on the Jake’s Math Lessons calculus 2 study guide, click here to go check that out and download your copy today.