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

[knowLittle]

Homework Statement

...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.

Homework Equations

The Attempt at a Solution

 

[DryRun]

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.

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

 

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

 

[LCKurtz]

Homework Statement

...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.

Homework Equations

The Attempt at a Solution

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

 

[LCKurtz]

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

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

 

[DryRun]

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

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.

 

[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.

 

[knowLittle]

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

 

[LCKurtz]

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

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

 

[knowLittle]

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

Thanks :) I'll try dydx.

 

[DryRun]

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

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.

 

[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.

 

[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)

 

[DryRun]

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)

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

 

[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?

 

[DryRun]

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?

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

 

[knowLittle]

It's not in the first octant.

 

[knowLittle]

Check figure.

 

[LCKurtz]

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.

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

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

 

[knowLittle]

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

 

[LCKurtz]

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

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)

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

When you set it up in rectangular coordinates, I assume that upper limit for dx is a typo and you meant $\sqrt{4-y^2}$. And your polar setup looks correct. They should both give the same answer. But I have doubts about this problem without seeing the full and exact statement of the problem. The plane $z=x+y$ goes negative when $x<0$, so I'm not sure you are calculating what you are supposed to be calculating.

 

[knowLittle]

The whole statement is similar to what I already posted, here is the full statement:

3. Set up a double integral to find the volume of the solid bounded by the graphs of the equations.
a.) z=xy, z=0, y=x, x=1, first octant

b.) z=x+y, x^2 + y^2=4, y>=0

c.) x^2 +z^2=1, y^2+z^2= 1, first octant

 

[LCKurtz]

The whole statement is similar to what I already posted, here is the full statement:

3. Set up a double integral to find the volume of the solid bounded by the graphs of the equations.
a.) z=xy, z=0, y=x, x=1, first octant

b.) z=x+y, x^2 + y^2=4, y>=0

c.) x^2 +z^2=1, y^2+z^2= 1, first octant

As stated, there is no bottom surface for part b.

 

[knowLittle]

So, it can't be solved for lack of information.

Let me try part c. Thank you.