1. ## 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.

2. ## 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...

3. ## 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

4. ## 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=...

5. ## 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

6. ## 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?

7. ## 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 $z= 4- y^2$.

8. ## Re: Double integrals question.

wont x go from 0 to 3?

9. ## 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).

10. ## Re: Double integrals question.

Originally Posted by bllnsr
wont x go from 0 to 3?
Yes, I wrote that wrong.

11. ## 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.

12. ## Re: Double integrals question.

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.

13. ## Re: Double integrals question.

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:

Graph tilted to the right:

same graph tilted to the left:

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.