Set up integral

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January 23rd 2013, 06:50 PM
sdsu619

Set up integral

Hey I am trying to do this problem:
Setup an integral to find the volume for the region in the first octant that touches the
xy plane and is bounded by the plane that passes through the points (1, 2, 0), (0, 0, 1)
and (0, 0, 0) and the plane that passes through the points (1, 2, 0), (0, 0, 2) and (3, 0, 0).
You do not need to integrate.

I wasn't sure what to do so I found the equation of the two planes. For the first plane(first 3 points) I got y = 2x(Is this equation for a plane possible since there is a point at (0,0,1)? Then for the second I got z = -(2/3)x - (2/3)y.
I'm not sure what to do now though as I cant see how to find the bounds for all three variables in order to set up the triple integral.
Any help would be appreciated
January 23rd 2013, 07:32 PM
Prove It

Re: Set up integral

I agree that finding the equations of the planes is a good idea. Remembering that a plane of the form $\displaystyle \begin{align*} ax + by + cz = d \end{align*}$ has the same coefficients as its normal vector, we can get the normal vector by finding two vectors that lie on the plane and taking their cross product.

Two vectors that lie on the first plane are $\displaystyle \begin{align*} \mathbf{u}_1 = <1 - 0, 2 - 0, 0 - 0> = <1, 2, 0> \end{align*}$ and $\displaystyle \begin{align*} \mathbf{v}_1 = <0 - 0, 0 - 0, 1 - 0> = <0, 0, 1> \end{align*}$ . So the first normal vector is found by

$\displaystyle \begin{align*} \mathbf{n}_1 &= \mathbf{u}_1 \times \mathbf{v}_1 \\ &= \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{matrix} \right| \\ &= 0 \left| \begin{matrix} \mathbf{j} & \mathbf{k} \\ 2 & 0 \end{matrix} \right| - 0 \left| \begin{matrix} \mathbf{i} & \mathbf{k} \\ 1 & 0 \end{matrix} \right| + 1 \left| \begin{matrix} \mathbf{i} & \mathbf{j} \\ 1 & 2 \end{matrix} \right| \\ &= 2\mathbf{i} - \mathbf{j} \\ &= <2, -1, 0> \end{align*}$

So your first plane will have equation $\displaystyle \begin{align*} 2x - y + 0z = d \end{align*}$

Substituting in any of the points, such as (0, 0, 0), finds that $\displaystyle \begin{align*} d = 0 \end{align*}$ and so the equation of your plane is $\displaystyle \begin{align*} 2x - y + 0z = 0 \end{align*}$ , which as you already determined can be written as $\displaystyle \begin{align*} y = 2x \end{align*}$ . So really this is a line in the x-y plane, that is then shifted vertically indefinitely in every direction. So as long as you have any two points which satisfy $\displaystyle \begin{align*} y = 2x \end{align*}$ , then any value of z will be fine. So yes, this equation of your plane is possible and (0, 0, 1) lies on it.

I disagree with your calculation of the second plane. You have the points (1, 2, 0), (0, 0, 2) and (3, 0, 0) lying on it.

Two vectors which lie on the plane are $\displaystyle \begin{align*} \mathbf{u}_2 = <1-0, 2-0, 0-2> = <1, 2, -2> \end{align*}$ and $\displaystyle \begin{align*} \mathbf{v}_2 = <3-0, 0-0, 0-2> = <3, 0, -2> \end{align*}$ . So the normal vector is given by

$\displaystyle \begin{align*} \mathbf{n}_2 &= \mathbf{u}_2 \times \mathbf{v}_2 \\ &= \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \phantom{-}\mathbf{k} \\ 1 & 2 & -2 \\ 3 & 0 & -2 \end{matrix} \right| \\ &= 3 \left| \begin{matrix} \mathbf{j} & \phantom{-}\mathbf{k} \\ 2 & -2 \end{matrix} \right| - 0 \left| \begin{matrix} \mathbf{i} & \phantom{-}\mathbf{k} \\ 1 & -2 \end{matrix} \right| + (-2) \left| \begin{matrix} \mathbf{i} & \mathbf{j} \\ 1 & 2 \end{matrix} \right| \\ &= 3 \left( -2\mathbf{j} - 2\mathbf{k} \right) - 2 \left( 2\mathbf{i} - \mathbf{j} \right) \\ &= -6\mathbf{j} - 6\mathbf{k} - 4\mathbf{i} + 2\mathbf{j} \\ &= -4\mathbf{i} - 4\mathbf{j} - 6\mathbf{k} \\ &= <-4, -4, -6> \end{align*}$

So the equation of your plane will be $\displaystyle \begin{align*} -4x - 4y - 6z = d \end{align*}$ , and substituting a point on the plane, like (0, 0, 2), gives $\displaystyle \begin{align*} d = -12 \end{align*}$ . So the equation of your plane is

$\displaystyle \begin{align*} -4x - 4y - 6z &= -12 \\ 2x + 2y + 3z &= 6 \\ 3z &= 6 - 2x - 2y \\ z &= 2 - \frac{2}{3}x - \frac{2}{3}y \end{align*}$

Now it would be a good idea to draw your region, so that you can decide which variables would be easiest to integrate over first.
January 23rd 2013, 08:02 PM
sdsu619

Re: Set up integral

Thanks for your help. I drew the region in the xy plane and came up with, in order of dzdydx: z(from 0 to 2-(2/3)x-(2/3)y) y(from 3-x to 2x) and x(from 0 to 1). Can anyone verify this?
January 24th 2013, 02:58 AM
tom@ballooncalculus

Re: Set up integral

Quote:

Originally Posted by sdsu619

Thanks for your help. I drew the region in the xy plane and came up with, in order of dzdydx: z(from 0 to 2-(2/3)x-(2/3)y) y(from 3-x to 2x) and x(from 0 to 1). Can anyone verify this?

Your z limits are correct for collecting the volume under the 'roof' which is z = 2 - (2/3)x - (2/3)y. But surely the 'floor' is a triangle going between 0 and 3 in the x direction...? If you do choose x as a first direction of travel, then you'll need to split the area of the floor in two separate integrals...

$\int_0^1\ 2x\ dx\ \equiv\ \int_0^1\ \int_{0}^{2x}\ 1\ dy\ dx$

and

$\int_1^3\ 3 - x\ dx\ \equiv\ \int_1^3\ \int_0^{3 - x}\ 1\ dy\ dx$

Which is fine, but you may prefer to travel y-wards first, between 0 and 2, during which you can integrate the area of the floor in one go, from strips of height x = [3 - y] - [(1/2)y] = 3 - (3/2)y...

$\int_0^2\ 3 - \tfrac{3}{2}y\ dy\ \equiv\int_0^2\ \int_0^{3 - \tfrac{3}{2}y}\ 1\ dx\ dy$
January 24th 2013, 09:26 AM
sdsu619

Re: Set up integral

Ah ok i get it I was looking at the wrong region. thanks for the input