Determine the Volume of the solid ?

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Homework Statement



determine the volume of the solid S which is bounded above by the surface z=10-x^2 and below by the z=Y^2+1;


Homework Equations





The Attempt at a Solution



what i did was first i tried to find the intersecting points of two surfaces .So i equal two functions and i got x^2+y^2=9 which is a circle center (0,0) and radius 3.then i used polar coordinates to evaluate this .

(r goes from 0 to 3 and theta goes from 0 to 2*pi )integrate (9-x^2-y^2)drd@.

am i correct ??or have i mad any kind of error?please explain?
 
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rclakmal said:

Homework Statement



determine the volume of the solid S which is bounded above by the surface z=10-x^2 and below by the z=Y^2+1;


Homework Equations





The Attempt at a Solution



what i did was first i tried to find the intersecting points of two surfaces .So i equal two functions and i got x^2+y^2=9 which is a circle center (0,0) and radius 3.then i used polar coordinates to evaluate this .

(r goes from 0 to 3 and theta goes from 0 to 2*pi )integrate (9-x^2-y^2)drd@.

am i correct ??or have i mad any kind of error?please explain?

Usually you have the function (of course that needs to be in polar coordinates) not only with dr dtheta, but r dr dtheta, so from what you have done the integral is the function 9r - r^3 with respect to r and then theta. Someone should confirm that though since I haven't looked at this in a while.
 
ah yr i forgot it thnaks !but am i right up to that point ?about setting the limits of the integral and other process ?
 
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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