## Merciless Group Title Let R be the planar region between the curves y^2 + (x-1)^2 = 4 and y^2 = 3(x-1) which contains the point (0, 0). 1.) Calculate the total area of ​​R. 2.) Determine the total length of the boundary of R. one year ago one year ago

1. Merciless Group Title

Here's the plot, x going from -1 to 3, y from -(squareroot(3)) to squareroot(3)

2. UnkleRhaukus Group Title

|dw:1354694787271:dw|

3. Merciless Group Title

??? Isn't R the region in between?

4. Merciless Group Title

|dw:1354694904341:dw|

5. UnkleRhaukus Group Title

|dw:1354694942463:dw|

6. Merciless Group Title

I still have no idea how to calculate the total area or boundary

7. UnkleRhaukus Group Title

so the region is the area of a circle , minus that parabola bit

8. Merciless Group Title

makes sense

9. UnkleRhaukus Group Title

can you find the radius of the circle

10. Merciless Group Title

11. UnkleRhaukus Group Title

so the area of the circle is ...

12. Merciless Group Title

4pi

13. UnkleRhaukus Group Title

right

14. Merciless Group Title

and then?

15. UnkleRhaukus Group Title

um,

16. UnkleRhaukus Group Title

maybe some integration

17. Merciless Group Title

that's exactly what I need, would love to learn how

18. UnkleRhaukus Group Title

have you found the intersection points ?

19. Merciless Group Title

intersections at (2, squareroot(3)), and (2, -squareroot(3))

20. UnkleRhaukus Group Title

|dw:1354696097101:dw|

21. UnkleRhaukus Group Title

lets calculate this area |dw:1354696370978:dw|

22. UnkleRhaukus Group Title

we can find the area of this region by integrating |dw:1354696620853:dw| $=\int\limits_{x=1}^{x=2}\int\limits_{y=0}^{y=\sqrt{3(x-1)}}\text dy\text dx$ $=\int\limits_{y=0}^{y=\sqrt{3}}\int\limits_{x=\tfrac{y^2}3+1 }^{x=2}\text dx\text dy$

23. UnkleRhaukus Group Title

maybe there is any easier way im not seeing

24. Merciless Group Title

hmm

25. amistre64 Group Title

youre doing ok, you just seem to be cutting it up way to much

26. amistre64 Group Title

y^2 + (x-1)^2 = 4 and y^2 = 3(x-1) move it so that the circle is centered at the origin; the area stay the same, we just move it y^2 + (x-1+1)^2 = 4 and y^2 = 3(x-1+1) y^2 + x^2 = 4 and y^2 = 3x x = sqrt(2^2-y^2) and x = 1/3 y^2 and it really doesnt matter if we have this oreiented long the x or y soo if its simpler to play with as y = sqrt(2^2-x^2) and y = 1/3 x^2 thats fine too

27. amistre64 Group Title

so, what does all this moving and rewrite do? |dw:1354733679539:dw| it just makes it easier to see the equation to integrate with

28. amistre64 Group Title

the area to remove is between the circle and parabola $\int_{a}^{b}\sqrt{4-x^2}-\frac13x^2~dx$

29. Merciless Group Title

and what is a,b?

30. amistre64 Group Title

the intersections

31. amistre64 Group Title

+- sqrt(3)

32. Merciless Group Title

Thanks! What about the total length of the boundary of R?

33. amistre64 Group Title

hardest part might be in determining the length of the parabolas arc, circles are a dime a dozen

34. amistre64 Group Title

do you recall any details about how to work a line integral?

35. amistre64 Group Title

|dw:1354757262089:dw|

36. amistre64 Group Title

$\int_{C}ds=\int_{a}^{b}\sqrt{(x')^2+(y' )^2}$

37. amistre64 Group Title

since x = x x' = 1 y = 1/3 x^2 y' = 2/3 x $\int_{a}^{b}\sqrt{(1)^2+(\frac23x )^2}~dx$

38. Merciless Group Title

that integral equals the boundary?

39. amistre64 Group Title

the integral equals the length of the curve of the parabola im sure you can figure out the circle part

40. amistre64 Group Title

|dw:1354758192335:dw|

41. amistre64 Group Title

find that and times 2 for the circle part

42. Merciless Group Title

|dw:1354758715223:dw|

43. Merciless Group Title

so 2x 1 + the integral of the curve (4.4) and I'd have 6.4 as total boundary length?

44. amistre64 Group Title

$y=\sqrt{4-x^2}$ $y'=\frac{-x}{\sqrt{4-x^2}}$ $\int_{-2}^{\sqrt3}\sqrt{1+\frac{x^2}{4-x^2}}dx$

45. amistre64 Group Title

circle part is, already did the parabola $2\int_{-2}^{\sqrt3}\sqrt{1+\frac{x^2}{4-x^2}}dx$ $2\int_{-2}^{\sqrt3}\sqrt{\frac{4}{4-x^2}}dx$ $2\int_{-2}^{\sqrt3}\frac{2}{\sqrt{{4-x^2}}}dx$ $dx=2sin(u)~:~dx=2cos(u) du$ $2\int_{-2}^{\sqrt3}\frac{2}{\sqrt{{4-4sin^2(u)}}}2cos(u) du$ $2\int_{-2}^{\sqrt3}\frac{2}{2\sqrt{{1-sin^2(u)}}}2cos(u) du$ $2\int_{-2}^{\sqrt3}\frac{2}{2cos^2(u)}2cos(u) du$ $2\int_{-2}^{\sqrt3}\frac{2}{cos(u)} du$ $4\int_{-2}^{\sqrt3}sec(u) du$ $4(ln(sec(u)+tan(u))$and either adjust back to xs, or adujst the limits in terms of us