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 [itex]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'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't seem to figure out what I am doing wrong. Any help is appreciated.[/itex]

 

[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.