Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Merciless

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

  • This Question is Closed
  1. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
    1 Attachment
  2. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1354694787271:dw|

    • one year ago
  3. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  4. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1354694904341:dw|

    • one year ago
  5. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1354694942463:dw|

    • one year ago
  6. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  7. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  8. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    makes sense

    • one year ago
  9. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    can you find the radius of the circle

    • one year ago
  10. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    the radius is 2

    • one year ago
  11. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    so the area of the circle is ...

    • one year ago
  12. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    4pi

    • one year ago
  13. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    right

    • one year ago
  14. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    and then?

    • one year ago
  15. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    um,

    • one year ago
  16. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    maybe some integration

    • one year ago
  17. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  18. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    have you found the intersection points ?

    • one year ago
  19. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  20. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1354696097101:dw|

    • one year ago
  21. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    lets calculate this area |dw:1354696370978:dw|

    • one year ago
  22. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  23. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 2

    maybe there is any easier way im not seeing

    • one year ago
  24. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm

    • one year ago
  25. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  26. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  27. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  28. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  29. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    and what is a,b?

    • one year ago
  30. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    the intersections

    • one year ago
  31. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    +- sqrt(3)

    • one year ago
  32. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  33. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  34. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  35. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1354757262089:dw|

    • one year ago
  36. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  37. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  38. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    that integral equals the boundary?

    • one year ago
  39. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  40. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1354758192335:dw|

    • one year ago
  41. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    find that and times 2 for the circle part

    • one year ago
  42. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1354758715223:dw|

    • one year ago
  43. Merciless
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  44. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  45. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.