anonymous one year ago Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis for y=x y=0 and x=2

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1. Zale101

First, draw the graphs and identify the enclosed area. Have you done that already?

2. anonymous

Yes

3. Zale101

|dw:1434058993462:dw|

4. Zale101

Since we are revolving the shell around the y-axis, we will be using the example i circled. |dw:1434059042077:dw|

5. Zale101

Makes sense?

6. anonymous

yeah! except your axis kind of confuse me I thought the y and x were switched

7. anonymous

But yeah I understand.

8. Zale101

But before anything else. You will analyze the rectangle and see what function along the x-axis is higher and what functions where the x-axis is lower. |dw:1434059232453:dw|

9. anonymous

ok I think I get it. The one thing im really confused on is the integral. do you use the numbers on the x-axis for the integral?

10. anonymous

$\int\limits_{0}^{3} 2\pi (x ^{2})$ so like this ?

11. anonymous

i used that and then plugged the integral in so i got $18\pi$ as my answer is that right. I think we may have learned it a different way because we never used that Big X little x thing we use a set equation that combines circumference and height

12. Zale101

|dw:1434059661157:dw|

13. Zale101

I forgot the mention, the intervals are based on where the spot is enclosed on the y-axis because we have dy and everything should be based on Y.

14. Zale101

First step: Let's identify the intervals. What do you think the intervals are?

15. anonymous

i gave you the wrong one by accident the problem says y=0 to x=3 so i think those are the intervals

16. anonymous

i told you 2 but its 3 sorry about that

17. Zale101

okay

18. Zale101

So the intervals were right. But what did you get for the big x and the small x?

19. anonymous

I don't know because the formula we learned dose not have that anywhere in there so i don't understand that.

20. anonymous

I honestly thought i had it right and just wanted to double check my answers and make sure but now im super confused

21. Zale101

$$\large a(x)=2\pi(radius)(height)$$

22. anonymous

Yeah!

23. Zale101

|dw:1434060538490:dw| $$\large a(x)=2pi(y)(X-x)$$

24. Zale101

$\Large \int\limits_{0}^{3}2 \pi y(X-x)dy$

25. anonymous

ok i still don't get the X-x thing i thought it was just the equation which is in this case just x but then u times that by x from 2pi move the 2 pi out front because its a constant take the anti derivative of the x^2 so then you end up with $2\pi \int\limits_{0}^{3} 1/3 x^{3}$

26. Zale101

|dw:1434060977099:dw| The radius is y, and the height is (X-x)

27. anonymous

thanks for helping btw i know its a pain

28. Zale101

|dw:1434061121877:dw| The big X is x=3 and the small x is y=x

29. Zale101

Makes sense? Hope I didn't confused you...

30. anonymous

ok yeah that makes sense.

31. anonymous

ok! so then the next step is the anti derivative of that?

32. Zale101

Yes.

33. Zale101

Use the fundamental theorem of calculus to find the definite integral. $\Large \int\limits_{a}^{b}f(y)dy=F(b)-F(a)$

34. anonymous

so 9pi or am i still not getting it?

35. anonymous

omg! ok thank you so much!!

36. Zale101

Np!

37. courtneygraley009

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