# Solving for the Volume of a Solid Using Double Integrals

• Temp0
The issue was with my bounds for the double integral. Thanks for checking my work!In summary, the volume of the solid bounded above by the surface z = x^2 + y^2 and below by the triangular region in the xy-plane is 1024/3. The method used was to find the height and base of the solid and set up a double integral to solve for the volume. After correcting the bounds for the double integral, the correct answer was found to be 2048/3.
Temp0

## Homework Statement

Find the volume of the solid bounded above by the surface z = x^2 + y^2 and below by the
triangular region in the xy-plane enclosed by the lines x = 0 , y = x , and x + y = 8.

V = ∫∫ Height
Base

## The Attempt at a Solution

I first found height, because height = z (upper) - z (lower) = (x^2 + y^2) - 0 = x^2 + y^2

Afterwards, I began solving for the base. I know the base is enclosed by the curves y = x,
y = x - 8, and x = 0. I proceed assuming the region is y-simple, giving me the boundaries:
Base { x ≤ y ≤ 8-x
and 0 ≤ x ≤ 4 }
This sets up the double integral for me, which turns out to be:
4 8-x
∫ { ∫ (x^2 + y^2) dy } dx
0 x

Solving this integral gives me that volume is 1024/3.
Could someone look through my work and see if I made any errors? I'm a beginner at setting up these equations, so I suspect my integral to have a mistake. Thank you in advance.

Temp0 said:

## Homework Statement

Find the volume of the solid bounded above by the surface z = x^2 + y^2 and below by the
triangular region in the xy-plane enclosed by the lines x = 0 , y = x , and x + y = 8.

V = ∫∫ Height
Base

## The Attempt at a Solution

I first found height, because height = z (upper) - z (lower) = (x^2 + y^2) - 0 = x^2 + y^2

Afterwards, I began solving for the base. I know the base is enclosed by the curves y = x,
y = x - 8, and x = 0. I proceed assuming the region is y-simple, giving me the boundaries:
Base { x ≤ y ≤ 8-x
and 0 ≤ x ≤ 4 }
This sets up the double integral for me, which turns out to be:
4 8-x
∫ { ∫ (x^2 + y^2) dy } dx
0 x

Solving this integral gives me that volume is 1024/3.
Could someone look through my work and see if I made any errors? I'm a beginner at setting up these equations, so I suspect my integral to have a mistake. Thank you in advance.

The way I managed to get that was by starting with ##x = y## and ##x = 8 - y##.

Then ##x = 0## implies that ##y = 0## and ##y = 8##.

Temp0 said:

## Homework Statement

Find the volume of the solid bounded above by the surface z = x^2 + y^2 and below by the
triangular region in the xy-plane enclosed by the lines x = 0 , y = x , and x + y = 8.

V = ∫∫ Height
Base

## The Attempt at a Solution

I first found height, because height = z (upper) - z (lower) = (x^2 + y^2) - 0 = x^2 + y^2

Afterwards, I began solving for the base. I know the base is enclosed by the curves y = x,
y = x - 8, and x = 0. I proceed assuming the region is y-simple, giving me the boundaries:
Base { x ≤ y ≤ 8-x
and 0 ≤ x ≤ 4 }
This sets up the double integral for me, which turns out to be:
4 8-x
∫ { ∫ (x^2 + y^2) dy } dx
0 x

Solving this integral gives me that volume is 1024/3.
Could someone look through my work and see if I made any errors? I'm a beginner at setting up these equations, so I suspect my integral to have a mistake. Thank you in advance.

Yup, thank you, I got the correct answer.

## 1. What is the concept of "Solving for the Volume of a Solid Using Double Integrals"?

Solving for the Volume of a Solid Using Double Integrals is a mathematical technique used in calculus to find the volume of a three-dimensional object. It involves using a double integral, which is an integral with two variables, to calculate the volume of a solid shape.

## 2. How do you set up a double integral for finding the volume of a solid?

To set up a double integral for finding the volume of a solid, you first need to determine the limits of integration for both variables. This involves understanding the boundaries of the solid in terms of x and y coordinates. Then, you can write the integral in the form of ∫∫f(x,y)dA, where f(x,y) represents the height of the solid at a given point and dA represents the infinitesimal area of a small region in the x-y plane.

## 3. What are the steps for solving a double integral to find the volume of a solid?

The steps for solving a double integral to find the volume of a solid are as follows: 1. Identify the limits of integration for both variables. 2. Set up the double integral in the form of ∫∫f(x,y)dA. 3. Evaluate the inner integral with respect to one variable while treating the other as a constant. 4. Evaluate the outer integral with respect to the remaining variable. 5. Simplify the expression and solve for the volume of the solid.

## 4. What types of shapes can be solved for using double integrals to find volume?

Double integrals can be used to find the volume of any three-dimensional shape, including prisms, cylinders, cones, spheres, and more complex shapes such as toroids and spheroids. As long as the boundaries of the shape can be defined in terms of x and y coordinates, a double integral can be set up to solve for its volume.

## 5. What are some real-world applications of solving for the volume of a solid using double integrals?

Solving for the volume of a solid using double integrals has many practical applications in fields such as architecture, engineering, and physics. For example, it can be used to calculate the volume of a building or bridge, determine the capacity of a water tank, or find the mass of a 3D object. It is also used in fluid mechanics to calculate the volume of a fluid flowing through a pipe or channel.

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