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anonymous

  • 5 years ago

how do you find the integral of the following: integral sqrt 9-h^2 limits 0 to 3

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  1. anonymous
    • 5 years ago
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    Are you learning about trig substitutions or polar coordinates?

  2. anonymous
    • 5 years ago
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    hey i have a test and i am review for it, its trig sunsitution

  3. anonymous
    • 5 years ago
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    substitute = 3 sin x

  4. anonymous
    • 5 years ago
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    Since in the limit of integral,0<h<3, so u can substitute h=3sin(a) since sin is btw 0 and 1, then (9-h^2) becomes 9cos^(a) take sqrt=3|cos(a)| also dh=d(3sin(a))=3cos(a)da thus and change limits h=0=3sin(a)=>a=0 h=3=3sin(a)=>a=pi/2 thus integral becomes limit 0 to 1 9cos^2(a)da put cos^2(a)=(2cos^(2a)-1)/2 then integrate

  5. anonymous
    • 5 years ago
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    Okay, since we have sqrt(a^2 - u^2) form, we have to use a sin t substitution. Let u = asint \[h = 3\sin \theta\] You should find that sine of theta is equal to h / 3. Use SOHCAHTOA to draw a right triangle. One leg should be h and the hypotenuse should be 3. Find the other leg using the pythagorean theorem to get sqrt(9 - h^2). Take the cosine of the angle \[\cos \theta=\sqrt {9-h^2}/3\] Multiply by 3 on both sides to get cosine(theta)/3 to use as a substitution. Substitute that for sqrt(9 - h^2) in the integral. Now go back to the h = 3sin(theta) equation and take the derivative to get dh = 3cos(theta). Substitute 3cos(theta) for dh to get the integral: \[\int\limits_{\theta(0)}^{\theta(3)}\cos \theta /3*3\cos \theta d \theta\] \[\int\limits_{\theta (0)}^{\theta (3)}\cos ^2 \theta d \theta\] You'll then have to integrate that and back-substitute.

  6. anonymous
    • 5 years ago
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    you can also draw a circle with radius 9 and conclude that the area is just one-fourth of the circle...

  7. anonymous
    • 5 years ago
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    radius 3, silly me

  8. anonymous
    • 5 years ago
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    okay thank you for the explanation but this question but this question is related to another problem u think u can help me

  9. anonymous
    • 5 years ago
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    ok

  10. anonymous
    • 5 years ago
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    can u please click on the link to see the picture i have to find the area by writing the definite integral and evaluate it http://www.twiddla.com/504415

  11. anonymous
    • 5 years ago
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    What area do you need? The area between the two chords?

  12. anonymous
    • 5 years ago
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    no thats the strip and u have to use the strip to find the area of the entire circle

  13. anonymous
    • 5 years ago
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    you would have to first write riemann sum then the definite integral

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