## Merciless 2 years ago 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.

1. Merciless

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

2. UnkleRhaukus

|dw:1354694787271:dw|

3. Merciless

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

4. Merciless

|dw:1354694904341:dw|

5. UnkleRhaukus

|dw:1354694942463:dw|

6. Merciless

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

7. UnkleRhaukus

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

8. Merciless

makes sense

9. UnkleRhaukus

can you find the radius of the circle

10. Merciless

11. UnkleRhaukus

so the area of the circle is ...

12. Merciless

4pi

13. UnkleRhaukus

right

14. Merciless

and then?

15. UnkleRhaukus

um,

16. UnkleRhaukus

maybe some integration

17. Merciless

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

18. UnkleRhaukus

have you found the intersection points ?

19. Merciless

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

20. UnkleRhaukus

|dw:1354696097101:dw|

21. UnkleRhaukus

lets calculate this area |dw:1354696370978:dw|

22. UnkleRhaukus

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

maybe there is any easier way im not seeing

24. Merciless

hmm

25. amistre64

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

26. amistre64

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

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

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

29. Merciless

and what is a,b?

30. amistre64

the intersections

31. amistre64

+- sqrt(3)

32. Merciless

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

33. amistre64

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

34. amistre64

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

35. amistre64

|dw:1354757262089:dw|

36. amistre64

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

37. amistre64

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

that integral equals the boundary?

39. amistre64

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

40. amistre64

|dw:1354758192335:dw|

41. amistre64

find that and times 2 for the circle part

42. Merciless

|dw:1354758715223:dw|

43. Merciless

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

44. amistre64

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

45. amistre64

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