# Set Up Double Integral to find Vol. Solid, Bounded by Graphs

1. Apr 26, 2012

### knowLittle

1. The problem statement, all variables and given/known data
...Bounded by graphs of equations:
z=xy,
z=0,
y=x,
x=1

I don't know what z=xy is. The rest of boundaries are clear.
I assume that when y=1 and x=1, z=1. But, is this a z=1 plane?
Check my figure attached.

Thank you.
2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### untitled.PNG
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2. Apr 26, 2012

### sharks

You can use a 3D plotter for the graph of z=xy. I have attached the graph.

Now, you can better visualize the required volume. It's an object with a triangular base and the top surface is z=xy.

$$\int^1_0 \int^y_1 xy\,.dxdy$$

#### Attached Files:

• ###### z=xy.gif
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3. Apr 26, 2012

### knowLittle

Yeah, I plotted it in Matlab as well, but I still couldn't draw it.
So, how can I build my double integral?

I found that if I horizontally sliced my the triangular base I get:
0<=y<= x
y<=x<=1
Now, I have no clue how to involve the upper bound. As sharks said, "the top surface".

Would it be $\int _{y}^{1}\int _{0}^{x}xy\left( dydx\right)$ ?

4. Apr 26, 2012

### LCKurtz

Are you missing a $y=0$ boundary? And, assuming that, Sharks has his limits wrong.

5. Apr 26, 2012

### LCKurtz

No. The outer integral can never have a variable limit if the result is to be a number.

6. Apr 26, 2012

### sharks

You are obviously correct.

knowLittle, a boundary (plane) is missing at the back of your drawing. If it's not been given, then ignore my suggestion.

7. Apr 26, 2012

### knowLittle

Sorry about that LC, It's restricted to 1st octant.
The full problem is:
Set up a double integral to find the volume of the solid bounded by the graphs of the equations.
z=xy,
z=0,
y=x,
x=1,
1st Octant.

8. Apr 26, 2012

### knowLittle

So, I guess that the equation would be:
$\int _{0}^{1}\int _{y}^{1}xydxdy$

9. Apr 26, 2012

### LCKurtz

Yes. It's even slightly less work to do dydx.

10. Apr 26, 2012

### knowLittle

Thanks :) I'll try dydx.

11. Apr 26, 2012

### sharks

Is that equation the same as what i suggested earlier?
$$\int^1_0 \int^y_1 xy\,.dxdy$$
The limits for when x varies are reversed. Does that mean, the volume will have an opposite sign?

EDIT: Indeed, it will have an opposite sign.

Last edited: Apr 26, 2012
12. Apr 26, 2012

### knowLittle

It gives the opposite volume, but I'm not sure if it's always valid.

I have another question from the same problem:
Part B:
z= x+y,
x^2+y^2=4,
y>=0

From the equation of the circle and Y>=0, the region limits are:
- (4-y^2)^(1/2) <=x <= (4-y^2) ^(1/2)
0<=y<=2

The volume equation:
$\int _{0}^{2}\int _{-\sqrt {4-y^{2}}}^{\sqrt {4-y^{2}}}\left( x+y\right) dxdy$
And, I got -(4/3) 2^(1/2) , but I don't think it's correct.

Last edited: Apr 26, 2012
13. Apr 26, 2012

### knowLittle

I tried a polar approach:

$\int _{0}^{\pi }\int _{0}^{2}\left( r\cos \theta + r\sin \theta \right) rdrd\theta$

Is this correct?
My result is :
(8/3) (sin (pi) - cos(pi) +1)

14. Apr 26, 2012

### sharks

Your limits for $\theta$ are not correct. I assume that part (b) requires the volume from the 1st octant?

15. Apr 26, 2012

### knowLittle

They do not mention 1st octant in part B. However, 1st octant is mentioned in part A and part C.

What's wrong with my limits for theta?

16. Apr 26, 2012

### sharks

If the volume is in the first octant, then θ varies between 0 and ∏/2

17. Apr 26, 2012

### knowLittle

It's not in the first octant.

18. Apr 26, 2012

### knowLittle

Check figure.

#### Attached Files:

• ###### untitled.PNG
File size:
1.4 KB
Views:
137
19. Apr 26, 2012

### LCKurtz

When you set up an area or volume integral, you must always do your integrations in the positive directions of the variables.

20. Apr 26, 2012

### knowLittle

Thank you, LC. Could you help me with part B of my problem?