Multiple Integration - Enclosed Volume

[Chaos2009]

Homework Statement

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

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 y before I began so I solved for when z = 0 which got me y = 0 and x = 0[/tex] which I then plugged into the first surface and found y = \pm\frac{\sqrt{3}}{2}. Well, z was only positive in the first and third quadrants, and I didn&#039;t see how a region could be enclosed in the third quadrant to I used y = 0 and y = \frac{\sqrt{3}}{2}.<br /> <br /> I then set up my integrals like so:<br /> <br /> \int_0^{\sqrt{3}/2} \!\!\! \int_0^{3-4y^2} x^3y \,dx \,dy<br /> <br /> Then integrated:<br /> <br /> \int_0^{\sqrt{3}/2} \frac{1}{4} \left[ x^4y \right]_{x=0}^{x=3-4y^2} \,dy<br /> <br /> \frac{1}{4} \int_0^{\sqrt{3}/2} y\left(3-4y^2\right)^4 \,dy<br /> <br /> -\frac{1}{160} \left[ \left(3-4y^2\right)^5 \right]_{y=0}^{y=\sqrt{3}/2}<br /> <br /> \frac{243}{160}<br /> <br /> However the back of the book says the answer is \frac{243}{80}, and I can&#039;t seem to figure out what I am doing wrong. Any help is appreciated.

 

[phyzguy]

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.