Solving for the Volume of a Solid Using Double Integrals

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.
  • #1
Temp0
79
0

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.


Homework Equations


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.
 
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  • #2
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.


Homework Equations


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.

Hmm I'm getting ##\frac{2048}{3}##. Double your answer.

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##.
 
  • #3
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.


Homework Equations


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.

Your setup and answer look correct to me.
 
  • #4
Yup, thank you, I got the correct answer.
 

Related to Solving for the Volume of a Solid Using Double Integrals

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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