Solving the Equation of a Sphere with Diameter Endpoints

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Homework Statement



Write the equation of the sphere whose diameter has endpoints at (-1,1,-3) and (3,-1,2).



Homework Equations





The Attempt at a Solution



(x-x1)2+ (y-y1)2+ (z-z1)2= r2

the center of the sphere is : (1,0,-1/2)
r= \sqrt{}5/4

then the equation is: (x-1)2+(y-0)2+(z+1/2)2=5/4

Please someone check my answer and if I am not right, help me to solve it.
 
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mamma_mia66 said:
the center of the sphere is : (1,0,-1/2)
r= \sqrt{}5/4

Hi mamma_mia66! :smile:

Centre is right. :biggrin:

But r is wrong. :cry:​

(you've calculated the distance from the centre to the origin! :wink:)
 
I am not sure that I know how to find the r ? too old to remember it after so many years not being in school.
 
Yes you do! :biggrin:

they've given you a diameter … :wink:​
 
r= sq. rt. (3-1)2+ (1-0)2+(2-1/2)2= 21/4
 
Last edited:

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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