Double Integrals for Volume in the First Octant: Iterated Integral Methods

[RaoulDuke]

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]

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]

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.