Is this graph schetched right?

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In the sollutions manual is this picture:
http://img7.imageshack.us/my.php?image=graph.png

But how can this be right? If you put x=0 or z=0, you are supposed to get hyperbolas in the yz-and xz-plane, but they are straight lines?

I think you will get a hyperbola in the yz-plane cause you get the equation y^2-16z^2=16, this can be written like (y^2)/4^2-z^2=1. Have I overlooked anything, or is the sollutions manual wrong here?
 
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hyper said:
In the sollutions manual is this picture:
http://img7.imageshack.us/my.php?image=graph.png

But how can this be right? If you put x=0 or z=0, you are supposed to get hyperbolas in the yz-and xz-plane, but they are straight lines?

I think you will get a hyperbola in the yz-plane cause you get the equation y^2-16z^2=16, this can be written like (y^2)/4^2-z^2=1. Have I overlooked anything, or is the sollutions manual wrong here?

I'm not sure what your question is. Yes, if x= 0 you get a hyperbola in the yz-plane and similarly for y= 0. I don't know why you say "but they are straight lines". If you are saying they look like straight lines in the picture, that picture simply is large enough or detailed enough to tell. The figure is a "hyperboloid of two sheets".
 
But look in the yz-plane, we are supposed to see the hyperbola y^2/(16)-z^2=1. The means it is supposed the cross the y-axis at y=+-4, but it goes all the way to the origin?
 
I take back what I said before. That is a hyperboloid of one sheet and there is a "hole" of radius 4 around the z-axis. Where did you get that picture? It looks like it might be the graph of 16z^2- x^2- y^2= 16[/math].
 
Thank you. I got it from the sollutions manual from Thomas Calculus.
 
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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