# Double integral volume of a region bounded by two z planes

• mahrap
In summary, the conversation discusses finding the volume of a region enclosed by two functions, z = 1 - y^2 and z = y^2 - 1, for x greater or equal to 0 and less than or equal to 2. The speaker suggests splitting the volume into two integrals and determining the bounds by sketching the region. They also discuss the importance of choosing the correct region and determining the range for y.
mahrap
a) find the volume of the region enclosed by
z = 1 - y^2 and z = y^2 -1 for x greater or equal to 0 and less than or equal to 2.

b) would i split up the volume into two integrals, each integral for each z function and then add them together? I also don't know how to find the bounds. Please help with thorough steps

mahrap said:
b) would i split up the volume into two integrals, each integral for each z function and then add them together?
The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?

haruspex said:
The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?

When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
(z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
( 1 - y^2 - (y^2 - 1) )

mahrap said:
When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
(z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
( 1 - y^2 - (y^2 - 1) )
You need a range for y which does not involve z.
Pls answer my question: do you think you want the region 1 - y2 < z < y2-1 or 1 - y2 > z > y2-1? Once you've chosen that you'll see what the range for y is.

## 1. What is a double integral?

A double integral is a mathematical concept used to calculate the volume of a three-dimensional region bounded by two z planes. It involves integrating a function over a two-dimensional region, with both x and y variables.

## 2. How is the volume of a region bounded by two z planes calculated using a double integral?

The volume of a region bounded by two z planes can be calculated by setting up a double integral with the boundaries defined by the two z planes and integrating the function over the two-dimensional region.

## 3. What is the importance of calculating the double integral volume of a region bounded by two z planes?

Calculating the double integral volume of a region bounded by two z planes is important in various fields of science, such as physics and engineering, as it allows for the determination of the volume of complex three-dimensional shapes and can be used to solve problems related to fluid mechanics, electromagnetism, and more.

## 4. Can the double integral volume of a region bounded by two z planes be negative?

Yes, the double integral volume of a region bounded by two z planes can be negative. This can occur when the function being integrated has negative values in certain regions of the two-dimensional region being integrated over.

## 5. Are there any limitations to using a double integral to calculate the volume of a region bounded by two z planes?

Yes, there are limitations to using a double integral to calculate the volume of a region bounded by two z planes. The shape of the region must be well-defined and the function being integrated must be continuous over the entire region. Additionally, the boundaries of the region must be known and defined in terms of the x and y variables.

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