Multiple Integration - Enclosed Volume

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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.
 

Answers and Replies

  • #2
phyzguy
Science Advisor
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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.
 

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