# Calculating volume using triple integrals

• kikifast4u
In summary, the student is trying to find the volume of a solid enclosed between a cylinder and two planes but is having trouble setting the integration limits.
kikifast4u

## Homework Statement

Find the volume of the solid enclosed between the cylinder x2+y2=9 and planes z=1 and x+z=5

V=∫∫∫dz dy dz

## The Attempt at a Solution

The problem I have here is setting the integration limits. I first tried using:
• z from 1 to 5-x
• y from √(9-x2) to -3
• x from -3 to 3

However, that gave me a negative answer, so I doubt that's the way to do it.

I then used polar coordinates for x and y and used the integral:
V=∫∫∫r dz dr dθ with limits
• z from 1 to 5-r*cos(θ)
• r from 0 to 3
• θ from 0 to 2∏

This time I got a positive answer, but I'm not sure whether the method is correct. We were never taught to use polar coordinates in volume integrals, so I'm not sure whether it's fine to mix them up.

I'm still struggling with this. Any help guys?

kikifast4u said:

## Homework Statement

Find the volume of the solid enclosed between the cylinder x2+y2=9 and planes z=1 and x+z=5

V=∫∫∫dz dy dz

## The Attempt at a Solution

The problem I have here is setting the integration limits. I first tried using:
• z from 1 to 5-x

• Okay, that's good.

[*]y from √(9-x2) to -3
Why -3? The cylinder x2+ y2= 9 goes from $-\sqrt{9- x^2}$ to $\sqrt{9- x^2}$.
[*]x from -3 to 3
However, that gave me a negative answer, so I doubt that's the way to do it.

I then used polar coordinates for x and y and used the integral:
V=∫∫∫r dz dr dθ with limits
• z from 1 to 5-r*cos(θ)
• r from 0 to 3
• θ from 0 to 2∏

This time I got a positive answer, but I'm not sure whether the method is correct. We were never taught to use polar coordinates in volume integrals, so I'm not sure whether it's fine to mix them up.

Thank you very much! I got the same answer as in the second method, so both are fine. Stupid stupid mistake!

## What is the formula for calculating volume using triple integrals?

The formula for calculating the volume using triple integrals is ∭fdV, where f is the function being integrated and dV represents the volume element.

## What is the difference between a single, double, and triple integral?

A single integral calculates the area under a curve in one dimension, a double integral calculates the volume under a surface in two dimensions, and a triple integral calculates the volume under a solid in three dimensions.

## How do you set up a triple integral to calculate volume?

To set up a triple integral for calculating volume, you need to identify the bounds for each variable, which will depend on the shape of the solid. The innermost integral will have the bounds for the z variable, the middle integral will have the bounds for the y variable, and the outermost integral will have the bounds for the x variable.

## What are some common mistakes when calculating volume using triple integrals?

Some common mistakes when calculating volume using triple integrals include forgetting to include the correct bounds for each variable, using the wrong order for the variables, and not properly setting up the integrals to match the shape of the solid.

## How can triple integrals be used in real-world applications?

Triple integrals are commonly used in physics and engineering to calculate the volume of objects with complex shapes, such as fluid flow in pipes or the mass of a three-dimensional object. They can also be used in economics to calculate the volume of production or consumption of goods and services.

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