anonymous 5 years ago Find the volume of the solid generated by revolving the region bound by the graphs of the equations around the line x=5. y=5-x, y=0, y=4, x=0

1. amistre64

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2. amistre64

id shell it

3. anonymous

What integral do I take for using discs though?

4. amistre64

disc is just using circles instead of cylindars;$\int pi f(x)^2dx$

5. amistre64

but it might be best to set it up better, our radius is in the form $$f^{-1}(x)$$ and there is an inner and outer radius to account for

6. amistre64

y=5-x x=5-y x-5=-y x+5=y would be our radius function: or y+5=x might be a better notation

7. amistre64

the outer radius is just "5" the inner radius is what we have to remove, right?

8. amistre64

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9. amistre64

\[piR^2-pir^2\=pi(R^2-r^2)=Area. of. ring/]

10. amistre64

\[pi R^2-pi r^2=pi(R^2-r^2)=Area. of. ring/]

11. amistre64

dint format on my screen hmmm

12. amistre64

since the radius is in terms of "x" values pi int(5^2 - (y+5)^2) dy, from 0 to 4 looks right