Rotating the Line y=-1 in Bounded Region R (y=9-x^2, y=0, x=0)

xstetsonx
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Consider the region R bounded by y=9-x^2, y=0, x=0. rotate the line y=-1

I am not sure about the bounds. The outer radius is -1 , and the inner radius is -10+x^2 right? but after i do the calculation i got a negative value. does that mean i got the radius wrong?
 
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Hi xstetsonx! :smile:

(try using the X2 tag just above the Reply box :wink:)
xstetsonx said:
Consider the region R bounded by y=9-x^2, y=0, x=0. rotate the line y=-1

I am not sure about the bounds. The outer radius is -1 , and the inner radius is -10+x^2 right? but after i do the calculation i got a negative value. does that mean i got the radius wrong?

How can a radius be negative? :confused:

The inner radius is 1 , and the outer radius is 10 - x2. :smile:
 
hahahaha omg i am an idiot hahahaha
 
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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