Evaluating a Double Integral: Finding the Volume of a Bounded Region

[stunner5000pt]

More fun yaaay
evaluate \int \int \int_{G} x^2 yz dx dy dz
where G is bounded by plane z=0, z=x, y=1, y=x

certrainly zi s bounded below by 0 and above by x. and y is boundedbelow by 1 and above by x. having a hard time picturing this...

i don't think this would pictured how the double integrals were, is there a way to visualize this?? Let x come out of the apge, y go right and z upwards
then z=0 is the x y plane, y =1 come out the page y=x come out of the and goes right and finally z=x come out the page and goes upward
but i don't think this helped determining the bounds did it??
Plase help

 

[xanthym]

More fun yaaay
evaluate \int \int \int_{G} x^2 yz dx dy dz
where G is bounded by plane z=0, z=x, y=1, y=x

certrainly zi s bounded below by 0 and above by x. and y is boundedbelow by 1 and above by x. having a hard time picturing this...

i don't think this would pictured how the double integrals were, is there a way to visualize this?? Let x come out of the apge, y go right and z upwards
then z=0 is the x y plane, y =1 come out the page y=x come out of the and goes right and finally z=x come out the page and goes upward
but i don't think this helped determining the bounds did it??
Plase help

SOLUTION HINTS:
Try the following:

1: \ \ \ \ \int_{x=0}^{1} \int_{y=1}^{x} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx \ \ =

2: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \ \left( \int_{z=0}^{x} z \, dz \right) \ dy \right ) \ dx \ \ =

3: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \ \left( \, \left[ z^{2}/2 \right]_{0}^{x} \, \right) \ dy \right ) \ dx \ \ =

4: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \cdot (x^{2}/2) \ dy \right ) \ dx \ \ =

5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=1}^{x} y \ dy \right ) \ dx \ \ =

Continue this process until all integrations have been performed and evaluated.


~~

 

[ehild]

evaluate \int \int \int_{G} x^2 yz dx dy dz
where G is bounded by plane z=0, z=x, y=1, y=x

SOLUTION HINTS:
Try the following:

1: \ \ \ \ \int_{x=0}^{1} \int_{y=1}^{x} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx
...


5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=1}^{x} y \ dy \right ) \ dx
...

Xanthym, the result is negative, that should not be. Change over the bound of integration with respect to y.

Stunner, you should sketch a diagram to find the bounds of integration. You have to find a closed volume as the domain of integration. That is a tetrahedron in this case, with vertexes (0,0,0), (1,1,1), (0,1,0) and (1,1,0) Look at the picture. So it is either

\int_{x=0}^{1} \int_{y=x}^{1} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx

or

\int_{y=0}^{1} \int_{x=0}^{y} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dx \, dy.

ehild

 

[xanthym]

(Bounds on y-integration corrected, now {y="x→1"} instead of previous {y="1→x"}, thanks to ehild's comment.)
SOLUTION HINTS:

1: \ \ \ \ \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx \ \ =

2: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \ \left( \int_{z=0}^{x} z \, dz \right) \ dy \right ) \ dx \ \ =

3: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \ \left( \, \left[ z^{2}/2 \right]_{0}^{x} \, \right) \ dy \right ) \ dx \ \ =

4: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \cdot (x^{2}/2) \ dy \right ) \ dx \ \ =

5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=x}^{1} y \ dy \right ) \ dx \ \ =

6: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left( \, \left[ y^{2}/2 \right]_{x}^{1} \, \right) \ dx \ \ =

Continue this process until all integrations have been performed and evaluated.


~~