## blackstreet23 one year ago What was the calculus formula to determine the volume of a solid? I don't mean solid of revolution. Just regular solid.

1. blackstreet23

volumes of cross sections

2. freckles

Are you sure you aren't looking for: $\text{ Volume }=\int\limits_a^bA(x) dx \text{ where } A(x) \text{ is the area of the cross section } ?$

3. blackstreet23

yes that is it ! thank you !

4. freckles

that's also the formula used in solids formed by revolution of some curve about some axis

5. LynFran

u forget to square it..i think @freckles

6. freckles

you know where we take circles as the cross section

7. freckles

no A(x) is the area you don't want to square the area

8. LynFran

o ok

9. freckles

if the cross section is a circle then A(x)=pi*r^2

10. freckles

where r is a function of x

11. freckles

though we could look at things in terms of y too not just x

12. freckles

you know depending on what suits us better

13. blackstreet23

but how is that the formula for solids of revolution? I mean the formula for solids of revolution is $\int\limits_{a}^{b}(\pi*f(x)^2) dx$

14. LynFran

i think the f(x) is actually the radius

15. blackstreet23

f(x) is the function

16. blackstreet23

i mean the height of the funtion

17. LynFran

yes which is intern the radius so A=pi*r^2

18. LynFran

and freckles just use the Area instead...i think

19. blackstreet23

I guess i am kind of confuse lol

20. LynFran

ok lets wait on till @freckles come online...

21. freckles

what's the question?

22. freckles

area of a circle is what I called A(x)

23. freckles

or the are of the cross section is A(x)

24. freckles

if the cross section is a circle then A(x) is pi*r^2 where r is the radius f(x) is the distance from the curve to the x-axis so f(x) is the radius if the axis of rotation is the x-axis

25. freckles

|dw:1440894543729:dw| Say we want to take this curve on [a,b] and rotate it about the x-axis

26. freckles

first reflecting the curve about the x-axis gives us: |dw:1440894612771:dw| now that drawing is not that great but that curve is basically suppose to be a mirror for the other curve I drew

27. freckles

|dw:1440894662165:dw| so we have a cross section that is a circle

28. freckles

notice the distance from the x-axis to the curve is indeed the radius which is called f(x) in this case

29. freckles

$\int\limits_{a}^{b}A(x) dx \\ \text{ so we have } A(x)=\pi \cdot (f(x))^2 \\$

30. LynFran

31. blackstreet23

ohh i see

32. freckles

|dw:1440894830121:dw| yes

33. freckles

but the A(x) is not squared

34. freckles

A(x) just means area of the cross section

35. blackstreet23

ohh i see ! I thought f(x) was just the height of the function

36. blackstreet23

It makes so much sense now :)

37. freckles

well you can also say |f(x)| is the height of the function

38. freckles

I put | | around it because f(x) could happen to be negative which just means the height is below the x-axis

39. LynFran

cool..