Volume in the First Octant

[Punkyc7]

Find the volume in the first octant bounded by the planes x+z=1 and y+2z=2. My question is where am I going wrong because when I use geometry I get the right answer but when I use calculus I do not.

So I solve for z and get
z=1-x
z=1-y/2

solve for y I get
y=2x

I set my first Double integral as
$$\int$$ from 0to2 $$\int$$ from 0toy/2 of 1-x dxdy

For my second integral I set up
$$\int$$ from 0to1 $$\int$$ 0to2x of1-y/2 dydx

solve those and I get 4/3 but the answer should be 2/3 because the volume is a Pyramid V=1/3 base area *height

 

[LCKurtz]

I suggest you draw a picture of the solid. Your inner limits are wrong on both integrals as you will see from a picture.

 

[Punkyc7]

I did draw a picture, that's how I knew it was a pyramid, so its the inner limits

 

[LCKurtz]

Perhaps it is just that you have your two integrands switched. Your regions don't coincide with the correct portions of the "roofs".

 

[Quinzio]

You basically have
a function
$$z=g(x)$$
and another
$$y=f(z)$$

the integral you search is:

$$\int_{0}^{x_0}\int_{0}^{g(x)}f(z)\ dz \ dx$$

where $$x_0$$ is the endpoint on the x axis.

If you solve it, you actually get 2/3.

I read you draw the solid, but you really carefully need to visualize, what you are integrating, in which variable, in which direction.
The inner integral sums towards z lines parallel to y.
The outer integral sums toward x areas parallel to the yz plane.
Writing all in neat Latex code really helps finding mistakes.
Hope it helps.