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Math Help - intergral by interpreting it in terms of areas

  1. #1
    Super Member bigwave's Avatar
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    intergral by interpreting it in terms of areas

    I am fairly new to this so...

    \int_{-3}^0(1+\sqrt{9- x^{2}})dx<br />
\Rightarrow<br />
\int_{-3}^{0}1dx + \int_{-3}^0\sqrt{9- x^{2}}dx<br />
\Rightarrow<br />
3 + \int_{-3}^0\sqrt{9- x^{2}}dx<br />

    assuming the form (from the reference in back of book)
     <br />
\int_{}{}\sqrt{a^2- u^2}du<br />
= \frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}{\frac{u}{a}+C<br />
    would be the way to solve the rest of it
    but don't seem to get the answer of 3+\frac{9\pi}{4}
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  2. #2
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    Quote Originally Posted by bigwave View Post
    I am fairly new to this so...

    \int_{-3}^0(1+\sqrt{9- x^{2}})dx<br />
\Rightarrow<br />
\int_{-3}^{0}1dx + \int_{-3}^0\sqrt{9- x^{2}}dx<br />
\Rightarrow<br />
3 + \int_{-3}^0\sqrt{9- x^{2}}dx<br />

    assuming the form (from the reference in back of book)
     <br />
\int_{}{}\sqrt{a^2- u^2}du<br />
= \frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}{\frac{u}{a}+C<br />
    would be the way to solve the rest of it
    but don't seem to get the answer of 3+\frac{9\pi}{4}
    I do. So clearly there's a mistake in your algebra. If you post all your working your mistake can be pointed out.
    Last edited by mr fantastic; June 28th 2010 at 08:57 PM.
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  3. #3
    Super Member bigwave's Avatar
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    so I am using the right form just have to carefull what I do with it... i will repost after i go thru this again.
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  4. #4
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    Quote Originally Posted by bigwave View Post
    so I am using the right form just have to carefull what I do with it... i will repost after i go thru this again.
    Drawing a picture always helps. From your post title, I assume you're finding the area of the circle (y - 1)^2 + x^2 = 3^2 that lies in the second quadrant of the xy-plane ....
    Last edited by mr fantastic; June 28th 2010 at 08:56 PM. Reason: Typo with quadrant.
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  5. #5
    Super Member bigwave's Avatar
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    think I plugged in the right values finally

    Quote Originally Posted by mr fantastic View Post
    Drawing a picture always helps. From your post title, I assume you're finding the area of the circle (y - 1)^2 + x^2 = 3^2 that lies in the third quadrant of the xy-plane ....
    actually I don't know how you came up with that, but not doubting it

    I think I plugged in the right values finally: (altho not sure about the x and C}


    <br />
\frac{0}{2}\sqrt{-3^2-0} + \frac{-3^2}{2}\sin^{-1}{\left(\frac{0}{9}\right)} + C<br />
\Rightarrow<br />
0+\frac{9}{2}\frac{\pi}{2} <br />
\Rightarrow<br />
9\frac{\pi}{4}<br />

    so the final answer would be

    3+9\frac{\pi}{4}

    btw I like the upgrades made to the MHF forum
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  6. #6
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    Quote Originally Posted by bigwave View Post
    actually I don't know how you came up with that, but not doubting it

    [snip]
    y = 1 + \sqrt{9 - x^2} \Rightarrow y - 1 = \sqrt{9 - x^2} \Rightarrow (y - 1)^2 = 9 - x^2 etc.

    Quote Originally Posted by bigwave View Post
    [snip]
    I think I plugged in the right values finally: (altho not sure about the x and C}


    <br />
\frac{0}{2}\sqrt{-3^2-0} + \frac{-3^2}{2}\sin^{-1}{\left(\frac{0}{9}\right)} + C<br />
\Rightarrow<br />
0+\frac{9}{2}\frac{\pi}{2} <br />
\Rightarrow<br />
9\frac{\pi}{4}<br />

    so the final answer would be

    3+9\frac{\pi}{4}

    btw I like the upgrades made to the MHF forum
    Sorry, but this working does not make a lot of sense to me. I expected to see an \sin^{-1} (-1) somewhere .... I also suggest you put brackets around negative numbers when squaring them (-3^2 is likely to be confused with -9 ....) Note also that the arbitrary constant is not required when finding definite integrals.
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  7. #7
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    mr fantastic is answering the question you asked but it is the wrong question! You appear to have completely misunderstood this problem.

    You titled this thread "Integral by interpreting it in terms of area" and you have made no attempt to that. The point of this problem is NOT to do any integration at all!

    If y= 1+ \sqrt{9- x^2}, then y- 1= \sqrt{9- x^2}, (y- 1)^2= 9- x^2 or x^2+ (y- 1)^2= 9. That's a circle with center at (0, 1) and radius 3. Because the original equation used only the positive square root, y= 1+ \sqrt{9- x^2}> 1 so we really have the upper half disk. Finally, because the integral is from x=-3 to 0, while the entire circle goes from x=-3 to x= +3, this integral is really the area of a quarter circle plus the rectangular section between that and the x-axis.

    What is the area of a circle of radius 3? What is the area of a quarter of that circle? What is the area of a rectangle of length 3 and height 1? What is the sum of those two areas?
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  8. #8
    Super Member bigwave's Avatar
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    thanks everyone for your help... just new to all this...
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