Integrate over the region enclosed by z = 1 + x + y and the xy plane

In summary, the problem involves finding the triple integral of E6xy over a region under the plane z = 1 + x + y. The limits of integration are -1≤x≤0, -1-x≤y≤0, and 0≤z≤ 1+x+y. To find the limits for y and x, you would plug in 0 for z and solve for each variable, resulting in y = -1 - x and x = -1. The upper bound for both y and x is 0. It is important to consider the 3 intercepts of the plane and draw a picture to find the limits of integration.
  • #1
Cloudless
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1. ∫∫∫E6xy dV, where E lies under the plane z = 1 + x + y

Apparently according to a classmate, the limits are:

-1≤x≤0, -1-x≤y≤0, and 0≤z≤ 1+x+y

I know how to get z. But I am confused for y and x.

To solve the limits of y, you would plug in 0 for z, getting y = -1 - x. But why is the upper bound 0?

Same for x. I assume you plug in 0 for both y and z when solving for x, resulting in x = -1. I am not sure why the upper bound is 0.
 
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  • #2
Cloudless said:
1. ∫∫∫E6xy dV, where E lies under the plane z = 1 + x + y

Apparently according to a classmate, the limits are:

-1≤x≤0, -1-x≤y≤0, and 0≤z≤ 1+x+y

I know how to get z. But I am confused for y and x.

To solve the limits of y, you would plug in 0 for z, getting y = -1 - x. But why is the upper bound 0?

Same for x. I assume you plug in 0 for both y and z when solving for x, resulting in x = -1. I am not sure why the upper bound is 0.

As stated, the problem has no finite answer because as x and y increase, so does z and there is an infinite region "under" the plane. So the problem probably mentions that not just under the plane but probably limited by the coordinate planes. Assuming that, my advice to you is to draw a picture. Plot the 3 intercepts of the plane giving a triangular portion and look at the xy domain for that part of the plane. That's where you will find the limits.
 

FAQ: Integrate over the region enclosed by z = 1 + x + y and the xy plane

What is the equation for the region enclosed by z = 1 + x + y and the xy plane?

The equation for the region enclosed by z = 1 + x + y and the xy plane is simply z = 1 + x + y. This is because the xy plane is defined by the equation z = 0, so by setting z = 1 + x + y, the region becomes bounded by the plane and the given curve.

How do I find the limits of integration for this region?

The limits of integration for this region can be found by setting the equation z = 1 + x + y equal to 0 and solving for x and y. This will give you two equations that can be used to determine the boundaries of the region in the xy plane.

What is the purpose of integrating over this region?

The purpose of integrating over this region is to find the volume of the solid bounded by the given surface and the xy plane. By integrating the function z = 1 + x + y over this region, we can find the total amount of space enclosed by the surface.

Can this region be integrated using a double or triple integral?

Yes, this region can be integrated using either a double or triple integral. Since the region is bounded by a curve and a plane, a double integral can be used to find the volume of the solid. However, if the function z = 1 + x + y is multiplied by a third variable, then a triple integral would be necessary to find the enclosed volume.

What are some real-world applications of integrating over a region like this?

Integrating over a region like this can have applications in various fields such as physics, engineering, and economics. For example, in physics, this type of integration can be used to find the center of mass of a solid object. In engineering, it can be used to determine the volume of a complex shape. In economics, it can be used to calculate the total profit or cost of a business by integrating over a region of supply and demand curves.

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