Visualising solid for double integration.

[ProPatto16]

Homework Statement

find volume of solid bounded by planes z=x, y=x, x+y=2, z=0

im struggling to visualise the solid.

can anyone help me? I am not sure how so illustrate but any pointers would be appreciated!

thank you!

 

[tiny-tim]

Hi ProPatto16!

z=x, y=x, x+y=2, z=0

Start with the easy ones … y=x, x+y=2.

These are both vertical planes, at a right-angle to each other: one intersects the "ground" at the 45° line through the origin, the other at the minus 45° line, through … ?

Then z = x is a slope through the y axis, and of course z = 0 is the "ground".

 

[ProPatto16]

Yeah I actually went and got an ice cream container and squares of card and tried to make it haha. With the x-axis along the bottom and the y up the left side, I worked out all the planes but then there seems to be no top on the shape. And I need another vertical plane to bound the shape along the x axis?

I'm sorry this is so ambiguous. Once I get the shape I know I can do the question.

 

[ProPatto16]

Also I am advised to draw the xy plane showing the base of the shape, ie the area I'm integrating over. So in this case it would be the xy plane with lines y=x and y=2-x which gives a cross when x and y-axis go from 0-2 and the two lines intersect at 1,1. So which of those 4 triangles in that xy plane is the one I'm integrating over? The one on the left bounded by y? Or the one on the bottom bounded by x? The top one and right one can be omitted since not bounded by both curves.

 

[tiny-tim]

… I worked out all the planes but then there seems to be no top on the shape.

It has a sloping top, x = z

And I need another vertical plane to bound the shape along the x axis?

… So which of those 4 triangles in that xy plane is the one I'm integrating over? The one on the left bounded by y? Or the one on the bottom bounded by x? The top one and right one can be omitted since not bounded by both curves.

hmm …

you're right, aren't you?

eg every point (1,y,0.5) will be in the region, for any y < 1.

(i expect they meant to say "y = z = 0" … try it like that! )

 

[ProPatto16]

Okay I think I see it now. My piece of card for x=z was down the side of my shape lol!
Ill see how I go and may reply back again tomorrow.
thanks! :)

 

[ProPatto16]

okay so I've come to this. since z = x

take double integral of x.dydx with x<y<2-x and 0<x<1

but since I am integrating by y first, and there is no y, its just the integral of x.dx between x and 2-x which becomes the integral of (2-2x).dx between 0 and 1.

which is 2x-x² between 0 and 1... subbing in 1 gives 2-1=1

so the volume of the solid is 1 unit cubed?