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### Course: Multivariable calculus>Unit 4

Lesson 7: Triple integrals

# Triple integrals 3

Figuring out the boundaries of integration. Created by Sal Khan.

## Want to join the conversation?

• Towards the very end, Why isn't upper y boundary y= -2x + 6? I'm not sure why its just 6..
• Every time you integrate, you are essentially removing a dimension from your problem as you approach the answer. The first integral removes the z dimension so z=0 the rest of the way through because you integrated with respect to z, which is why you use the 2 dimensional projection in the xy plane to get the upper bound of x. Once you integrate with respect to x, you have finished with that dimension and x=0 from then on. Finally, now that z=0 and x=0, you can plug those into the original surface equation to get your upper y bound of 6. Then, you do your final one-dimensional integral to solve the problem.
• Can your x integral be from 0 to 3?
• Only if you integrate that last and you use y=6-2x as the upper bound for y. That boundary must be used at least once and your last integration must only have constants.
• sal you left me hangin in the end where is the next video??
• Why z=2 is that green triangle surface?
• SAL,
PLEASE do more on tripple integrals! I love your videos, but its a bummer you stop here with this trivial example. As someone else menioned in a comment years ago, can you continue into clindrical and sphereical coordinat systems?? Paraboloids, and frustrums, and more complex surfaces?
• Why do we put the variable bounds in the innermost integral?
• because if we use it in the ¿outermost integral? we will have in the final result a function, and we can´t have this, we need to have a fixed number...
• At , if we want the area between z = 2 and 2x + 3z + y = 6, why do we integrate from x = 3 - y/2 to x= 0? Wouldn't we integrate from x = 3 to x = 3 - y/2 since x = 3 - y/2 is the lower bound?
• Very good observation. I followed in the footsteps of those who solved this hideous problem, without realizing this, and now I know my solution is meaningless since the limits are mixed up. I would recommend mentioning this as an error in the video so that others may be able to avoid similar meaningless work.
Thank you for your attention to detail and precision!
• At , i don't understand why you used the x-y plane instead of the x-z plane. I am not very good at visualizing 3D diagrams, so could you please show me why?
• Because if your integration order takes care of Z first, i.e. dzdxdy, then once you find your Z limits in the first integral, then you are done with Z altogether, the next step to solve the triple integral is to project into the remaining variables' plane, in other words, project into the x-y plane. You would only project your solid into the x-z plane if you integrated with the order dydxdz or dydzdx, in other words, integrating y first.
• I Don't Get the boundaries you put for the z, x and Y ? i mean why don't we take boundaries for z to be from 0 to 2 and in x from 0 to 3 and y from 0 to 6 ? Im really confused
• Because that assumes that at every given point in your object, the lower Z height is 0, but as you can see, the lower Z heights vary, directly based on where you are in x and y, i.e. different values of x and y produce different values of Z. This relation between x, y, and z, is described by that Plane: 2x+3z+y=6.
Integrating with x from 0 to 3, y from 0 to 6, and z from 0 to 2 would give you the mass of a rectangular box, but as you can see, the object Sal drew is not a rectangular box. Instead, if we want the correct limits of integration, we need to find the height of that plane at every given x, and y value, in other words, we need to solve for Z in the plane equation 2X+3Z+Y=6, well using algebra, Z= (-2X -y)/3 +2. That equation is still the plane, but it's in terms of X and Y, so providing X and Y values yields a Z value. Essentially, that Plane is your lower bound in your integration limit (note that this would only be your integration limit if you were integrating Z first: dZdYdX or dZdXdY). The upper bound is the flat plane Z=2, because it's height does not depend on the X and Y values.
Hope this helps, I understand your struggle, this stuff used to confuse me as well.