- #1

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i got an answer of (-91/72) x pi x (sqrt(2)-sqrt(3))...am i right? Thanks!

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- Thread starter mister_okay
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- #1

- 20

- 0

i got an answer of (-91/72) x pi x (sqrt(2)-sqrt(3))...am i right? Thanks!

- #2

arildno

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Well, the answer ought to be:

[tex]\frac{6^{3}-5^{3}}{3}(\frac{\pi}{3}-\frac{\pi}{4})(\cos\frac{\pi}{6}-\cos\frac{\pi}{4})[/tex]

So, if your answer is just a simplification of this, then it is correct.

EDIT:

Yup, it is correct.

[tex]\frac{6^{3}-5^{3}}{3}(\frac{\pi}{3}-\frac{\pi}{4})(\cos\frac{\pi}{6}-\cos\frac{\pi}{4})[/tex]

So, if your answer is just a simplification of this, then it is correct.

EDIT:

Yup, it is correct.

Last edited:

- #3

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but one question..

shouldnt it be cos(pi/4)-cos(pi/6) instead of cos(pi/6)-cos(pi/4)?

- #4

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No, since:

[tex]\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} sin\phi d\phi = [-cos\phi]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -cos(\frac{\pi}{4}) - (-cos(\frac{\pi}{6})) = cos(\frac{\pi}{6}) - cos(\frac{\pi}{4})[/tex]

[tex]\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} sin\phi d\phi = [-cos\phi]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -cos(\frac{\pi}{4}) - (-cos(\frac{\pi}{6})) = cos(\frac{\pi}{6}) - cos(\frac{\pi}{4})[/tex]

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- #5

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oh ok...forgot about the negative. thanks!

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