Multiple Integration - Enclosed Volume

In summary, the problem involves finding the volume of a region bounded by planes and surfaces using multiple integration. The attempt at a solution involved finding the bounds for y and setting up the integrals. The integration was then carried out, leading to a final answer of 243/160. However, the back of the book states a different answer, and further investigation suggests that the book may be incorrect.
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Homework Statement



Find the volume of the region bounded by the planes [tex]x = 0[/tex] and [tex]z = 0[/tex] and the surfaces [tex]x = -4 y ^ 2 + 3[/tex], and [tex]z = x ^ 3 y[/tex].

Homework Equations



The problem is listed in the book in the section Multiple Integration - Application of the Double Integral.

The Attempt at a Solution



I figured I needed to find the bounds on [tex]y[/tex] before I began so I solved for when [tex]z = 0[/tex] which got me [tex]y = 0[/tex] and [itex]x = 0[/tex] which I then plugged into the first surface and found [tex]y = \pm\frac{\sqrt{3}}{2}[/tex]. Well, [tex]z[/tex] was only positive in the first and third quadrants, and I didn't see how a region could be enclosed in the third quadrant to I used [tex]y = 0[/tex] and [tex]y = \frac{\sqrt{3}}{2}[/tex].

I then set up my integrals like so:

[tex]\int_0^{\sqrt{3}/2} \!\!\! \int_0^{3-4y^2} x^3y \,dx \,dy[/tex]

Then integrated:

[tex]\int_0^{\sqrt{3}/2} \frac{1}{4} \left[ x^4y \right]_{x=0}^{x=3-4y^2} \,dy[/tex]

[tex]\frac{1}{4} \int_0^{\sqrt{3}/2} y\left(3-4y^2\right)^4 \,dy[/tex]

[tex]-\frac{1}{160} \left[ \left(3-4y^2\right)^5 \right]_{y=0}^{y=\sqrt{3}/2}[/tex]

[tex]\frac{243}{160}[/tex]

However the back of the book says the answer is [tex]\frac{243}{80}[/tex], and I can't seem to figure out what I am doing wrong. Any help is appreciated.
 
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  • #2
I got interested in this problem, and went through it is some detail, including building a Mathematica notebook to plot the surfaces and check the integration. I'm convinced that you're right and the book is wrong. Let me know if you learn more.
 

1. What is multiple integration?

Multiple integration is a mathematical concept that involves finding the enclosed volume of a three-dimensional shape by integrating over multiple variables. It is used to solve problems in physics, engineering, and other fields that involve complex, three-dimensional systems.

2. How is multiple integration used to find the enclosed volume of a shape?

Multiple integration is used to find the enclosed volume of a shape by dividing the shape into small, infinitesimal pieces and summing up the volume of each piece. This is done by integrating over the variables that determine the boundaries of each piece.

3. What are the different types of multiple integration?

There are two main types of multiple integration: double integration and triple integration. Double integration involves integrating over two variables, while triple integration involves integrating over three variables. Both types are used to find the enclosed volume of a three-dimensional shape.

4. What are some real-world applications of multiple integration?

Multiple integration has many real-world applications, such as calculating the volume of a solid object, finding the mass and center of mass of an object, determining the moment of inertia of an object, and solving problems in fluid mechanics and electromagnetism.

5. Are there any limitations or challenges to using multiple integration?

One of the main challenges of using multiple integration is setting up the integral correctly. This requires a good understanding of the shape and its boundaries. Additionally, multiple integration can become computationally intensive for complex shapes, making it difficult to find an exact solution. In these cases, numerical methods may be used to approximate the enclosed volume.

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