# Double integrals question.

• May 29th 2013, 08:34 PM
bllnsr
Double integrals question.
Find the volume of the solid in the first octant bounded by the coordinates planes, the plane x=3 and the parabolic cylinder z = 4-(y^2).

Can you please explain how to find limits of x and y in this question as I am not clear how to find them.
Help would be appreciated.
• May 29th 2013, 09:20 PM
Prove It
Re: Double integrals question.
Have you tried drawing a sketch of the region? Also, surely from the given information you can at least see the lower bound for each variable...
• May 29th 2013, 10:14 PM
bllnsr
Re: Double integrals question.
http://img196.imageshack.us/img196/6752/20130530162.jpg

here is what I sketched. I think the lower limit of x will be 0 and upper limit will be 3 but I don't know about y
• May 30th 2013, 12:26 AM
MINOANMAN
Re: Double integrals question.
Bills
Your figure is wrong. you are in 3D and the x=3 is not A LINE BUT A PLANE!!!
SEE THE FIGURE BELOW AND TRY TO DO SOME WORK...DO NOT FORGET THAT YOU MUST RESTRICT YOURSELF IN THE FIRST OCTANT.......the red is the plane x=3 the blue the surface z=...

Attachment 28489
• May 30th 2013, 09:28 AM
bllnsr
Re: Double integrals question.
Its to confusing for me I am having difficult time solving these questions.
here is what I sketched.

http://img818.imageshack.us/img818/2...0530211915.jpg
• May 30th 2013, 11:36 AM
bllnsr
Re: Double integrals question.
I have got a little bit now I think I have sketched it again.

http://img716.imageshack.us/img716/888/paraxa.jpg

the lower limit of y will be 0 but what about the upper limit?
• May 30th 2013, 12:20 PM
HallsofIvy
Re: Double integrals question.
Your graph is not very clear. I take it that parabola is supposed to be the cylinder " $z= 4- y^2$" but it is drawn in the wrong direction. It goes up to z= 4 , at y= 0 and crosses the y-axis, z= 0, at y= 2. And the cylindrical surface extends parallel to the x-axis.
It should be clear that x goes from 0 to 4, that for each x, y goes from 0 to 2, and that, for each x and y, z goes from 0 to $z= 4- y^2$.

Attachment 28492
• May 30th 2013, 03:07 PM
bllnsr
Re: Double integrals question.
wont x go from 0 to 3?
• May 31st 2013, 09:29 AM
mopen80
Re: Double integrals question.
Find first the area of the parabolic sector as int_0^2 (4-y^2)dy, then extrude it from x=0 to x=3. so V=(int_0^3(int_0^2 (4-y^2)dy)dx).
• May 31st 2013, 10:14 AM
HallsofIvy
Re: Double integrals question.
Quote:

Originally Posted by bllnsr
wont x go from 0 to 3?

Yes, I wrote that wrong.
• May 31st 2013, 10:18 AM
HallsofIvy
Re: Double integrals question.
The "footprint" in the xy- plane is a rectangle: x goes from 0 to 3 and y goes from 0 to 2.
• June 1st 2013, 01:29 AM
Prove It
Re: Double integrals question.
Quote:

Originally Posted by mopen80
Find first the area of the parabolic sector as int_0^2 (4-y^2)dy, then extrude it from x=0 to x=3. so V=(int_0^3(int_0^2 (4-y^2)dy)dx).

Of course, this only works because the solid formed is a prism, in other cases you will need to perform a double integral.
• June 2nd 2013, 01:29 PM
x3bnm
Re: Double integrals question.
Quote:

Originally Posted by bllnsr
Find the volume of the solid in the first octant bounded by the coordinates planes, the plane x=3 and the parabolic cylinder z = 4-(y^2).

Can you please explain how to find limits of x and y in this question as I am not clear how to find them.
Help would be appreciated.

Here's the full graph.
Graph from the side:

Attachment 28509

Graph tilted to the right:
Attachment 28508

same graph tilted to the left:
Attachment 28507

Here $z = 4 - y^2$ is cyan
$x = 3$ is dark blue
$x=0$ is red
$y = 0$ is gray
and $z = 0$ is yellow.

You need to find the volume of the area that is enclosed with cyan, dark blue, red, gray and yellow parts of the graph which is in the first octant.