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anonymous

  • one year ago

Evaluate the integral by interpreting it in terms of areas. S^0 on the bottom -6 (5+root36-x^2)dx I have no idea how to work this problem out, please help!

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  1. ganeshie8
    • one year ago
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    Take a good look at the integrand, does it look familiar to you ?

  2. anonymous
    • one year ago
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    the -6 is on the bottom of the integral, um no not really

  3. ganeshie8
    • one year ago
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    like this ?\[\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx\]

  4. anonymous
    • one year ago
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    yes

  5. ganeshie8
    • one year ago
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    May be rearrange the integrand a bit \[\large y=\color{blue}{5+\sqrt{36-x^2}}\] \[\large (y\color{blue}{-5})^2=(\color{blue}{\sqrt{36-x^2}})^2\] \[\large \color{blue}{x^2}+(y\color{blue}{-5})^2=\color{blue}{36}\] what about now ? seen this before ?

  6. anonymous
    • one year ago
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    it has to do with circles right?

  7. ganeshie8
    • one year ago
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    |dw:1436074459700:dw|

  8. ganeshie8
    • one year ago
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    look at only the top half, because the original function is \(y=5+\sqrt{36-x^2}\), which represents only the upper half of circle.

  9. anonymous
    • one year ago
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    okay

  10. ganeshie8
    • one year ago
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    you need to find the area under that curve between x=-6 and x=0

  11. ganeshie8
    • one year ago
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    |dw:1436074742051:dw|

  12. ganeshie8
    • one year ago
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    thats the area the given integral represents see if you can work it

  13. anonymous
    • one year ago
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    would my answer be 36pi/2? Im not really sure what Im supposed to do, this is the firts problem i have that is like this

  14. ganeshie8
    • one year ago
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    how did you get 36pi/2 ?

  15. anonymous
    • one year ago
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    1/2pir^2=1/2pi6^2=36pi/2 and then i did it another way and got 30

  16. ganeshie8
    • one year ago
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    nope, notice that the area is made up of a "quarter circle" and a "rectangle" : |dw:1436075697415:dw|

  17. ganeshie8
    • one year ago
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    find area of quarter circle, find area of rectangle add them up

  18. anonymous
    • one year ago
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    okay, what do i do (5+root36-x^2)

  19. anonymous
    • one year ago
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    would my answer be 6+9pi? @ganeshie8

  20. ganeshie8
    • one year ago
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    whats the area of rectangle ?

  21. ganeshie8
    • one year ago
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    |dw:1436076828225:dw|

  22. anonymous
    • one year ago
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    is that a 5? so it would be 25

  23. ganeshie8
    • one year ago
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    careful, you need to use area of rectangle formula not area of square

  24. anonymous
    • one year ago
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    w*l=6*6=36

  25. ganeshie8
    • one year ago
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    try again

  26. anonymous
    • one year ago
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    30

  27. ganeshie8
    • one year ago
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    Yes, save that, next find the area of quarter circle

  28. anonymous
    • one year ago
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    1/4PI*6^2=9pi

  29. anonymous
    • one year ago
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    @ganeshie8

  30. ganeshie8
    • one year ago
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    Looks good! add them up

  31. anonymous
    • one year ago
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    30+9pi

  32. ganeshie8
    • one year ago
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    \[\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx~~=6\times 5+\dfrac{\pi\times 6^2}{4} = 30+9\pi\]

  33. anonymous
    • one year ago
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    Thank you!!!

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