## 8867564 Group Title 1) Find the volume of the solid of revolution formed when the region bounded between x = 1, x = 4, y = x2 and y = 0 is revolved vertically around the x-axis. Using cans is the answer 21π? 2 years ago 2 years ago

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1. timo86m Group Title

by x2 you mean x*2 or x^2

2. timo86m Group Title

if x*2

3. timo86m Group Title

|dw:1337145858842:dw| I think

4. timo86m Group Title

area of a small slice will result in a ring so it be pi r2^2 pi r1^2

5. colorful Group Title

I know it as shell method never heard "cans" befor lol

6. colorful Group Title

|dw:1337146309981:dw|we need the height and radius of each cylinder...

7. colorful Group Title

|dw:1337146393852:dw|

8. colorful Group Title

the surface area of a cylinder is given by$2\pi rh$so in this case each surface area as a function of x is$A(x)=2\pi rh=2\pi x^2\cdot x=2\pi x^3$

9. colorful Group Title

adding up all these areas from x=1 to x=4 should give us the correct integral

10. colorful Group Title

$V=2\pi\int_{1}^{4}x^3dx$

11. 8867564 Group Title

Thank you for the help on this problem. It was very helpful.

12. timo86m Group Title

deos y=x^2 or x*2?

13. 8867564 Group Title

It's x^2

14. timo86m Group Title

oh

15. timo86m Group Title

you can take area from y=0 to y=1 use pi*4^2-pi*1^2=15 pi so that is first part

16. timo86m Group Title

now for 1 to 16 integral(pi*4^2-pi(sqrt(y))^2dy) from 1 to 16

17. timo86m Group Title

I got 127.5000000*Pi

18. timo86m Group Title

let me know what the answer was :)

19. marco26 Group Title