## anonymous 4 years ago *CALC2 HELP PLEASE* The region enclosed by the curve x=sqrt(y), and by the lines x= -y and y=5 rotated around the x-axis.

1. anonymous

Thanks for the medal, but do you have any tips?

2. anonymous

well, what you've got yourself there is an integral with some pretty odd bounds, give me a second so I can draw a picture to see how to approach it.

3. anonymous

Thanks man.. The shape is pretty funky..

4. anonymous

is it 3d?

5. anonymous

|dw:1327642062679:dw|

6. anonymous

yeah it's rotating

7. anonymous

another, question, which area contained by those lines?

8. anonymous

It doesn't specify really, so I'm confused there as well..

9. anonymous

The whole area I shaded in

10. anonymous

Because graph is this

11. anonymous

hmm.. that is really weird..

12. anonymous

aww, I thought dumbcow had it figured out.. haha

13. anonymous

so depending on what the area is its just a matter of doing the volume of revolution, but without knowing which specific area is... I want to assume that the area is the bit off to the left from y=-x up to y=5 and from y=-x over to y=x^2

14. anonymous

did they specify x=+sqrt(y)?

15. anonymous

ok i think you can approach this using the washer method where you have an outer area - inner area each cross section is a circle and you integrate wrt x from -5 to sqrt5 the outer radius would be 5 since that is the max y-value. the inner radius would be the y-value of the appropriate function f(x) |dw:1327642598764:dw| $\large V = \pi \int\limits_{-5}^{0}(25-x^{2}) dx + \pi \int\limits_{0}^{\sqrt{5}}(25-x^{4}) dx$

16. anonymous

dumbcow: after integration I got a negative volume?

17. anonymous

hmm no it should be positive make sure when you evaluate you subtract upper limit - lower limit

18. anonymous

ok let me try again

19. anonymous

the integration would be.. pi(25x-x^3/3) and pi(25x-x^5/5), right?

20. anonymous

yes

21. anonymous
22. anonymous

I got 402.296 on wolfram, and it says incorrect on my homework.. :/

23. anonymous

Nahhh I'm just kidding.. Worked perfect. Thanks guys!!

24. anonymous

isn't wolfram great. ok well do they want in terms of pi? i'll look at it again i may have made a mistake somewhere

25. anonymous

never mind :)