Write six different iterated triple integrals for the volume of the tetrahedron cut from the fir

Write six different iterated triple integrals for the volume...

Key Concepts

Order of Integration

The order of integration in an iterated integral determines the sequence in which the integrals are evaluated. Choosing different orders can lead to different limits of integration and may simplify the evaluation of the integral depending on the geometry of the region involved.

Triple Integral

A triple integral is an extension of a double integral to functions of three variables. It allows for the calculation of volumes, mass, and other quantities over three-dimensional regions by integrating the function in three sequential steps, each with its own variable of integration.

Iterated Integral

An iterated integral breaks down a triple integral into a series of integrals performed one after the other. The order of integration can be chosen to simplify the computation, and each integral is evaluated over limits that may depend on the remaining variables.

Volume of a Tetrahedron

Calculating the volume of a tetrahedron using integrals involves setting up a region defined by a plane and the coordinate planes. The tetrahedron in this context is a solid bounded by one slanted plane and the walls of the first octant, and its volume can be expressed as a triple integral over this region.

First Octant

The first octant refers to the portion of three-dimensional space where all three coordinates (x, y, and z) are non-negative. This condition restricts the limits of integration and defines one boundary of the region of integration.

Region of Integration

The region of integration is defined by the geometric boundaries of the problem, including the plane and the coordinate planes in the first octant. Setting up accurate variable limits that arise from these boundaries is crucial in computing the volume of the solid.

Transcript

00:02 So for this problem, we are looking for the volume of the tetrahedron cut by the first octant and this plane here, which is 6x plus 3y plus 2z equals 6.

00:22 So then the first octant is where x, y, and z are all positive.

00:29 So we can think of it as this octant here.

00:35 Okay, but that makes it a little hard to see, so i'm not going to leave that there.

00:40 Now we're using six different iterations to do this, right? because the volume is going to be a triple integral, so of this form.

00:56 But then the iterations have to do with the order of dv, which includes dz, d.

01:03 D.

01:03 Z, d .y, and dx.

01:07 So then we can order these differently, and that changes our bounds.

01:12 Okay, so i'm going to start with a pretty standard order, which is, dz, d, y, dx.

01:22 So starting with dz, then working your way to dy, and then to dx.

01:27 So, whatever this first integration is, we're going to take our equation here, and we're going to solve for z, because our first iteration is dz.

01:43 So i'm going to subtract by 6x and 3y to get 2z equals 6 minus 6x plus 3y.

01:53 Divide by 2 and we get z equals 3 minus 3x minus plus sorry 3 halves oh no that should be minus excuse that minus 3 halves y okay so that is our upper bound for our z integration and then our lower bound is zero right because the first octant means where all the values are positive okay so then we take this equation and we use it to solve for our y integration.

02:36 So we take this, we set z equal to zero, because that's already integrated, and we solve for y.

02:44 So we're left with 3 minus 3x minus 3 halves y.

02:52 Now we need equal zero.

02:54 Excuse me.

02:55 Add 3 halves y to both sides, and we get 3 minus 3x equals 3 halves y.

03:03 Then we want to divide by 3 halves, which is the same as multiple.

03:11 Xx3rds, and we're left with y equals 2 minus 2x.

03:19 And that gives us the upper bounds for our y integration.

03:22 So 2 minus 2x with a lower bound again of 0 because we're in the first octet.

03:32 And then lastly, we need our x integration.

03:39 We can take this equation here, set it equal to 0, y equal to 0 like we did before, and solve for x.

03:52 Add 2x to both sides and we get 2x equals 2.

03:57 So we have x equals 1.

03:59 And that's our upper bound for the x integration.

04:03 And our lower bound is 0 again because of the first octave.

04:08 Okay, so that's our first iteration.

04:13 And then we can easily move into a second iteration, which also begins with dz, but we're going to switch dx and d y.

04:27 So our first bound is still the same because our z integration comes first.

04:32 We have 3 minus 3x minus 3 halves y.

04:37 Okay, but then we go back over to our work over here and we want to go back to this step where we set z equal to 0 and instead of solving for y, we're going to solve for x.

04:54 All right, so add 3x to both sides and we're left with 3 minus 3 halves y.

05:04 Then divide by 3 and we have x and we have x and equals 1 minus y divided by 2.

05:14 And that is our upper bounds for our x integration on this iteration.

05:21 And then finally set the equate, set x equal to 0, and solve for y, for our y integration in this order.

05:32 Add y divided by 2 to both sides, multiply by 2, and we get y equals 2.

05:38 So that's our upper bound for the y iteration.

05:44 And again, the lower bound is zero because of the first octet.

05:51 Okay.

05:53 So with these two, to continue on, we have to start from scratch because our first integration is no longer going to be z.

06:05 Okay.

06:07 So now i'm going to start with an integration of y.

06:15 And then may as well just do d, y, d z, dx.

06:22 Okay.

06:23 That means we take our initial equation and we solve for y.

06:28 So i'm going to subtract 6x and 2z.

06:32 So we get 3y equals 6 minus 6x minus 2z.

06:41 Divide by 3.

06:42 We get y equals 2 minus 2x minus 2x minus 2.

06:48 2 thirds z.

06:53 And that is our upper bound for the y integration.

06:59 2 minus 2x minus 2 thirds z.

07:04 And our lower bound is 0 as we've been seeing.

07:08 Okay, now we set our equation, set z equal to 0, and because, sorry, set y equal to 0, i'm still trying to keep my mind in that first problem.

07:22 Which is not what we want.

07:25 And we solve for z because z is our second integration.

07:29 So 0 equals 2 minus 2x minus 2 thirds z.

07:37 Add 2 3 to both sides and we're left with 2 minus 2x.

07:45 I'm going to multiply by 3 first.

07:49 So 2 z equals 6 minus 6x and then divide by 2 and we're left with z equals z equals 3 minus 3x.

07:59 All right, so that's our upper bound for this second iteration.

08:08 And then lastly, set z equal to 0 and solve for x.

08:18 Add 3x to both sides.

08:21 We're left with x equals 1.

08:26 Okay, so we have three iterations now.

08:31 The next one i'm going to address is again starting with the iteration of y, right? so the integration of y, excuse me...

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