# Some double integrals

## Homework Statement

Set up the following as a double integral whose value is the stated volume, express this double in two ways as an iterated integral, and evaluate one of these.

## Homework Equations

Volume, in the first octant, bounded by -
z = 4- (y^2)
x=0
y=0
z=0
3x + 4y =12

## The Attempt at a Solution

Forgive me if I don't know my way completely around conveying the notation online.

y = (12 - 3x)/4

When y = 0, x = 4.

x = (12-4y)/3

when x = 0, y = 3.

I assume that I set up the integral in two ways, one where dA = dy dx and another where dA = dx dy.
$$\int$$$$^{4}_{0}$$ $$\int^{(12-3x)/4}_{0}$$ (4 - y$$^{2}$$) dy dx

or

$$\int$$$$^{3}_{0}$$ $$\int^{(12-4y)/3}_{0}$$ (4 - y$$^{2}$$) dx dy

and solve? Basically, did I set up the integrals correctly?

## Answers and Replies

Mark44
Mentor
First I'll fix up your LaTeX, and will then take another look a little later. What you did is not too bad, so kudos for getting as much of it right as you did. Tip: use only one pair of [ tex] [ /tex] tags, with everything inside the single pair. That works better than having a whole multitude of tex tags. You can click the integral to see what I did.

## Homework Statement

Set up the following as a double integral whose value is the stated volume, express this double in two ways as an iterated integral, and evaluate one of these.

## Homework Equations

Volume, in the first octant, bounded by -
z = 4- (y^2)
x=0
y=0
z=0
3x + 4y =12

## The Attempt at a Solution

Forgive me if I don't know my way completely around conveying the notation online.

y = (12 - 3x)/4

When y = 0, x = 4.

x = (12-4y)/3

when x = 0, y = 3.

I assume that I set up the integral in two ways, one where dA = dy dx and another where dA = dx dy.
$$\int^{4}_{0}\int^{(12-3x)/4}_{0} (4 - y^{2}) dy dx$$

or

$$\int^{3}_{0} \int^{(12-4y)/3}_{0} (4 - y^{2}) dx dy$$

and solve? Basically, did I set up the integrals correctly?

Mark44
Mentor
The second integral is close, but the first one is off by a lot, since two separate integrals are required and you have only one.

Setting up these integrals and changing from one order of integration to another requires a good understanding of the two-dimensional region over which integration takes place. Your mistake is in thinking the the region of integration is a triangle. It isn't; it's a trapezoid in the x-y plane bounded by the origin, and the points (4, 0, 0), (0, 2, 0), and the intersection point in the x-y plane of the plane 3x + 4y = 12 and the cylinder z = 4 - y^2. I didn't calculate its exact position, but it looks to be at about (1, 2, 0).

If you have strips that are parallel to the x-axis, the individual strips run from x = 0 to x = 4 - (4/3)y, and these strips sweep across from y = 0 to y = 2. One integral suffices.

If your strips are parallel to the y-axis, some of the strips run from y = 0 to y = 2, but the others run from y = 0 to y = 3 - (3/4)x. Since the strips have different bounds, you need two different integrals when you integrate first with respect to y then with respect to x.