Review: This year we have only done single integrals, both indefinite and definite. To review, an integral indicates that we must find the area beneath and curve. To do this, we have to find the antiderivative of a given function. If we are working with an indefinite integral, we will end up with a function in terms of x or y (depending on whether we are taking the integral with respect to dx or dy) + C, a constant of integration. However, if we are using a definite integral, we will have certain values that we evaluate the antiderivative at. See below for a visual:
But what happens if we are no longer looking at flat surfaces?
Double Integrals:
When you graph a curve on the x-y plane, you use a single integral to find he area beneath that curve. But what happens when you add a third axis, (let’s call it z) and that 2D plane becomes a 3D space? And what does something like that even look like? See below...
So that’s kinda wonky. But let’s say z is a function of whatever x and y value you happen to be at on the 2D plane. In other words, z is a function in terms of variables x and y. Mathematically, we’ll refer to it as f(x,y). But what does that look like? Well, you know that 3D shapes have length, width, and height. So if you think of x and y as your length and width, z must be your height. What does that form? It forms a shape, and it can be any shape, depending on what your specific f(x,y) function looks like. But to get started understanding this, let’s draw an arbitrary f(x,y) function on our triple-axes below...
Hmm, nifty. Now what if we want to find the volume of the space enclosed between that shape and the x-y plane. Well, as you already know, a single integral is what you use to find the area under a curve. Therefore, a double integral can be used to find the volume under a shape! Let’s start by setting boundaries. On what intervals of x and y would we like to find the volume of this space? Let’s set a constant x and a constant y, as shown below...
Wow, that looks super great! We’ll call our constant x endpoint “a”, and our constant y endpoint “b”. Finding the volume of that whole space probably seems pretty intimidating, but we’re super-duper AP Calc students! We’ve conquered over 200 O’Brien opportunity days, literally! A real, full-blown AP test! A million online applets... we can handle a double integral. Let’s first think about how we find a single integral. We take a bunch of tiny slivers and add them up, the length being the y value (which changes depending on which one you chose, making it a function), and the width being an infinitely small change in x, otherwise known as dx. For a 3D shape, it’s the same idea. We want to find the 2D areas of a bunch of little slivers of our 3D shape, and add them together. See the illustration below! The sliver we’d like to find the area of is highlighted in purple, and its infinitely small thickness is highlighted in green...
Hmm, nifty. Now what if we want to find the volume of the space enclosed between that shape and the x-y plane. Well, as you already know, a single integral is what you use to find the area under a curve. Therefore, a double integral can be used to find the volume under a shape! Let’s start by setting boundaries. On what intervals of x and y would we like to find the volume of this space? Let’s set a constant x and a constant y, as shown below...
Wow, that looks super great! We’ll call our constant x endpoint “a”, and our constant y endpoint “b”. Finding the volume of that whole space probably seems pretty intimidating, but we’re super-duper AP Calc students! We’ve conquered over 200 O’Brien opportunity days, literally! A real, full-blown AP test! A million online applets... we can handle a double integral. Let’s first think about how we find a single integral. We take a bunch of tiny slivers and add them up, the length being the y value (which changes depending on which one you chose, making it a function), and the width being an infinitely small change in x, otherwise known as dx. For a 3D shape, it’s the same idea. We want to find the 2D areas of a bunch of little slivers of our 3D shape, and add them together. See the illustration below! The sliver we’d like to find the area of is highlighted in purple, and its infinitely small thickness is highlighted in green...
Represented algebraically, that sliver looks like this: 
So what is the area of that 2D sliver? Well, it’s the length times height. We know the length is going to be x = a, and z is a function of x and y, remember? f(x,y). And remember to multiply by that infinitely small change in x for the width, dx.
Now you want to add a bunch of those slivers up--every sliver on the y interval from 0 to b. This is where the second integral comes in. You want to integrate that first integral, which was in terms of dx, again, in terms of dy. See below...
One very important thing to remember is that when you integrate your first function, f(x,y), you integrate it in terms of dx. This means you treat all y values as constants and only apply the power rule (in reverse, of course) to x terms. Then, the second time you integrate, you are integrating in terms of dy and therefore can apply the power rule to the remaining y terms, as all x terms have become constant through the FTC. Here, let’s do a practice problem:
So in this case, f(x,y) = x^2 + y^2. First, we need to integrate in terms of dx, so we treat the y term as a constant:
Now, just use the FTC to turn the x values into constants.
Ok, now it’s pretty simple: just integrate from 0 to b in terms of y and plug in your endpoints with the FTC for the final answer.
TA-DAAA!
First, we take a very small cube from our large cube and call its dimensions dxdydz or dV for short.
We then make three integrals regarding the dimensions of the x axis, the y axis, and the z axis, respectively:
Now we need to create a triple integral in order to find the volume. Because no density function is given, we can just assume the density of the pyramid is 1kg/m^3. We know with a triple integral, we will be summing up the small volume dxdydz within the proper bounds. It really doesn't matter what order we sum, so let's start with dz.
dz: We know that one of the planes will be z=0. The other one will be the surface given in terms of x and y: z = 6 - x - 2y. This means that the first part of our integral looks like such:
dy: We know that one of our y-axis limits will be y=0. If we look at the graph of the x-y trace, we see that it's also bounded by that diagonal line. Because the x-y trace does not involve z, we merely eliminate z from the given equation and solve for an equation in terms of x. The result is another integral:
Triple integrals seem to be commonly used to find the mass of a non-uniform massed object. The functions used are often in terms of x, y, and z and are representative of a variable density. We set these up the same way as before. Take the surface from above: z = 6 - x - 2y. Let's say they give us the variable density: p(x,y,z)=6-x. Then they ask us to find the mass. Because we already figured out the boundaries of the different axes, we can just use the triple integral we used earlier but instead of the density being 1, the density is now 6-x.
This is how far most problems will ask you to go because it becomes super complicated and shit. But we all know how to integrate so IN THEORY we could do it.
So what is the area of that 2D sliver? Well, it’s the length times height. We know the length is going to be x = a, and z is a function of x and y, remember? f(x,y). And remember to multiply by that infinitely small change in x for the width, dx.
Now you want to add a bunch of those slivers up--every sliver on the y interval from 0 to b. This is where the second integral comes in. You want to integrate that first integral, which was in terms of dx, again, in terms of dy. See below...
One very important thing to remember is that when you integrate your first function, f(x,y), you integrate it in terms of dx. This means you treat all y values as constants and only apply the power rule (in reverse, of course) to x terms. Then, the second time you integrate, you are integrating in terms of dy and therefore can apply the power rule to the remaining y terms, as all x terms have become constant through the FTC. Here, let’s do a practice problem:
So in this case, f(x,y) = x^2 + y^2. First, we need to integrate in terms of dx, so we treat the y term as a constant:
Now, just use the FTC to turn the x values into constants.
Ok, now it’s pretty simple: just integrate from 0 to b in terms of y and plug in your endpoints with the FTC for the final answer.
TA-DAAA!
Triple Integrals: Triple integrals involve functions that are in terms of x, y, and z.
One way a triple integral can be used is to find the volume of a box.
Let 0 ≤ x ≤ 5
Let 0 ≤ y ≤ 3
Let 0 ≤ y ≤ 3
Let 0 ≤ z ≤ 2
When these limits are graphed on an x, y, z plane, a cube is formed. To find the volume of this cube, we can just use basic geometry : L x W x H = (5)(3)(2)=30m^3. However, we can also use a triple integral to find the volume.
First, we take a very small cube from our large cube and call its dimensions dxdydz or dV for short.
We then make three integrals regarding the dimensions of the x axis, the y axis, and the z axis, respectively:
Note that there is a constant (1) in the triple integral above. This represents the density of the block which is 1 kg/m^3. Then we integrate as such:
Although this gave us the volume of the block (we can check our answer with geometry), it also gave us the mass. #gotlucky We know that the formula for density is d=m/v. In our integral, we are taking density and multiplying it by volume. Therefore, our answer should be mass. AND IT IS!!! If the density is 1 kg/m^3, the mass and the volume will be the same! YAY!
But how would our answer change if the density was 2 kg/m^3? Mass and volume would no longer be the same. The answer to the triple integral will now be the mass:
However, if the mass is 60 kg and the density is 2 kg/m^3, with the equation d=m/v, we find that our volume is still the same (as we would expect it to be - the cube's dimensions did not change).
One way we can use triple integrals is to find the volume bounded by certain curves. Let's say we have the curve x + 2y + z = 6, x=0, y=0, z=0. The surface itself looks like so:
But how would our answer change if the density was 2 kg/m^3? Mass and volume would no longer be the same. The answer to the triple integral will now be the mass:
If density isn't given, and the question asks for volume, just assume the density is 1 kg/m^3 or whatever unit is being used. It's just a rule of thumb.
This is just a mini intro....
One way we can use triple integrals is to find the volume bounded by certain curves. Let's say we have the curve x + 2y + z = 6, x=0, y=0, z=0. The surface itself looks like so:
(a semi-birds eye view; flat up-down is z-axis, flat left-right is x-axis, vertical up-down is z-axis)
How do we find the boundaries??? Well....
For the x-axis bounds, we know that x = 0. But for the upper bound? If we make y=0 and z=0 in the original equation x + 2y + z = 6, x=6. Therefore, for our graph, x=[0,6]. If we do this for y and z (making x and z equal 0 and making y and x equal 0, respectively), y = [0,3] and z = [0,6]. One thing that we can do is draw traces in order to see the what the three possible planes look like:
Now we need to create a triple integral in order to find the volume. Because no density function is given, we can just assume the density of the pyramid is 1kg/m^3. We know with a triple integral, we will be summing up the small volume dxdydz within the proper bounds. It really doesn't matter what order we sum, so let's start with dz.
dz: We know that one of the planes will be z=0. The other one will be the surface given in terms of x and y: z = 6 - x - 2y. This means that the first part of our integral looks like such:
dy: We know that one of our y-axis limits will be y=0. If we look at the graph of the x-y trace, we see that it's also bounded by that diagonal line. Because the x-y trace does not involve z, we merely eliminate z from the given equation and solve for an equation in terms of x. The result is another integral:
dx: The last limit is easy because we just sum up dx from 0 to 6. Straightforward. Our final product is:
Then we integrate as we did with the double integrals, treating the variables which we are NOT integrating with respect to as constants. Here is a picture from the online solution:
Triple integrals seem to be commonly used to find the mass of a non-uniform massed object. The functions used are often in terms of x, y, and z and are representative of a variable density. We set these up the same way as before. Take the surface from above: z = 6 - x - 2y. Let's say they give us the variable density: p(x,y,z)=6-x. Then they ask us to find the mass. Because we already figured out the boundaries of the different axes, we can just use the triple integral we used earlier but instead of the density being 1, the density is now 6-x.
#SENIORSUMMERBITCHEZZZ
Works Cited:
"Triple Integral Problems." Triple Integral Problems. N.p., n.d. Web. 22 May 2013. <http://homepages.math.uic.edu/~dcabrera/practice_exams/topics/tripleintegrals.html>.
"Double and Triple Integrals." Khan Academy. N.p., n.d. Web. 22 May 2013. <https://www.khanacademy.org/math/calculus/double_triple_integrals>.
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