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

xstetsonx
Messages
78
Reaction score
0
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?
 
Physics news on Phys.org
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top