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Merciless

  • 3 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.

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  1. Merciless
    • 3 years ago
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    Here's the plot, x going from -1 to 3, y from -(squareroot(3)) to squareroot(3)

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  2. UnkleRhaukus
    • 3 years ago
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    |dw:1354694787271:dw|

  3. Merciless
    • 3 years ago
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    ??? Isn't R the region in between?

  4. Merciless
    • 3 years ago
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    |dw:1354694904341:dw|

  5. UnkleRhaukus
    • 3 years ago
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    |dw:1354694942463:dw|

  6. Merciless
    • 3 years ago
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    I still have no idea how to calculate the total area or boundary

  7. UnkleRhaukus
    • 3 years ago
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    so the region is the area of a circle , minus that parabola bit

  8. Merciless
    • 3 years ago
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    makes sense

  9. UnkleRhaukus
    • 3 years ago
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    can you find the radius of the circle

  10. Merciless
    • 3 years ago
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    the radius is 2

  11. UnkleRhaukus
    • 3 years ago
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    so the area of the circle is ...

  12. Merciless
    • 3 years ago
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    4pi

  13. UnkleRhaukus
    • 3 years ago
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    right

  14. Merciless
    • 3 years ago
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    and then?

  15. UnkleRhaukus
    • 3 years ago
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    um,

  16. UnkleRhaukus
    • 3 years ago
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    maybe some integration

  17. Merciless
    • 3 years ago
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    that's exactly what I need, would love to learn how

  18. UnkleRhaukus
    • 3 years ago
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    have you found the intersection points ?

  19. Merciless
    • 3 years ago
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    intersections at (2, squareroot(3)), and (2, -squareroot(3))

  20. UnkleRhaukus
    • 3 years ago
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    |dw:1354696097101:dw|

  21. UnkleRhaukus
    • 3 years ago
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    lets calculate this area |dw:1354696370978:dw|

  22. UnkleRhaukus
    • 3 years ago
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    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
    • 3 years ago
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    maybe there is any easier way im not seeing

  24. Merciless
    • 3 years ago
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    hmm

  25. amistre64
    • 3 years ago
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    youre doing ok, you just seem to be cutting it up way to much

  26. amistre64
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    the area to remove is between the circle and parabola \[\int_{a}^{b}\sqrt{4-x^2}-\frac13x^2~dx\]

  29. Merciless
    • 3 years ago
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    and what is a,b?

  30. amistre64
    • 3 years ago
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    the intersections

  31. amistre64
    • 3 years ago
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    +- sqrt(3)

  32. Merciless
    • 3 years ago
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    Thanks! What about the total length of the boundary of R?

  33. amistre64
    • 3 years ago
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    hardest part might be in determining the length of the parabolas arc, circles are a dime a dozen

  34. amistre64
    • 3 years ago
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    do you recall any details about how to work a line integral?

  35. amistre64
    • 3 years ago
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    |dw:1354757262089:dw|

  36. amistre64
    • 3 years ago
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    \[\int_{C}ds=\int_{a}^{b}\sqrt{(x')^2+(y' )^2}\]

  37. amistre64
    • 3 years ago
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    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
    • 3 years ago
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    that integral equals the boundary?

  39. amistre64
    • 3 years ago
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    the integral equals the length of the curve of the parabola im sure you can figure out the circle part

  40. amistre64
    • 3 years ago
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    |dw:1354758192335:dw|

  41. amistre64
    • 3 years ago
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    find that and times 2 for the circle part

  42. Merciless
    • 3 years ago
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    |dw:1354758715223:dw|

  43. Merciless
    • 3 years ago
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    so 2x 1 + the integral of the curve (4.4) and I'd have 6.4 as total boundary length?

  44. amistre64
    • 3 years ago
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    \[y=\sqrt{4-x^2}\] \[y'=\frac{-x}{\sqrt{4-x^2}}\] \[\int_{-2}^{\sqrt3}\sqrt{1+\frac{x^2}{4-x^2}}dx\]

  45. amistre64
    • 3 years ago
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    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

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