Double (Iterated) Integration. Where am I going wrong?

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Homework Statement



Let D be the region in the x,y-plane enclosed by the curves x = (y^2) - 9 and x = 2y - 6. Find the volume of the solid with base D that lies under the surface z = 1y^2.

Homework Equations



Fubini's Theorem and Fundamental Theorem of Calculus

The Attempt at a Solution



Hello everyone,

I need some help with a simple iterated integration. I just can't seem to figure out where I am going wrong. According to an online homework site, my answer is not correct.

I am getting a crazy answer of -179975.2.

Below is the image of my work. Any help would be GREATLY appreciated, as this assignment is due soon.

I apologies for the large image for those with small screens.

[PLAIN]http://img139.imageshack.us/img139/1324/68070772.jpg
 
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Your region D is drawn incorrectly. The line intersects the curve at (-8, -1) and (0, 3). You have intersection points of (0, 3), which is mislabeled, and (16, 5). Your point at (0, 3) is below the x-axis, so should be labeled (0, -3).
 
Thank you. The only thing I had to change was the limit of 16 to -9 to 3 to -1 and got the answer correct with the same approach as above. You are correct about the fact that I drew the D region incorrectly... stupid mistake on my part.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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