anonymous
  • anonymous
Prove the circumference of a circle is 2πr @LeibyStrauss
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
The formula for a cirle is x^2 + y^2 = r^2 To solve for y: - x^2 from both sides
anonymous
  • anonymous
\[y^2 = r^2 - x^2\] take the square root of both sides \[y = \sqrt{r^2 - x^2}\]
anonymous
  • anonymous
integrate

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anonymous
  • anonymous
If we want to take the are of the 1st quadrant the limits (bounds) will be 0 - r The area of the entire circle is x 4 To find the area we can integrate: \[4 \int\limits_{0}^{r} \sqrt{r^2 - x^2}\] (forgot to put dx in the equation)
anonymous
  • anonymous
correct
anonymous
  • anonymous
okay i see
anonymous
  • anonymous
Now we can use substitution \[x = (r)\sin \theta\] if we differentiate both sides \[dx = (r)\cos \theta d \theta\] I will leave the bounds (limits) out for a moment \[4 \int\limits \sqrt{r^2 - [(r)\sin \theta]^2} (r)\cos \theta\]
anonymous
  • anonymous
\[x = rsin\]
anonymous
  • anonymous
the original bounds were 0 to r. If we set (r)sin theta to 0 \[0 = (r)\sin \theta\] divide both sides by r \[0 = \sin \theta\] \[\theta = 0\] because sin(0) = 0 Next we need to set (r)sin theta = r
anonymous
  • anonymous
\[r = (r) \sin \theta\] divide both sides by r \[1 = \sin \theta\] \[\theta = \frac{ \pi }{ 2 }\] because \[\sin (\frac{ \pi }{ 2 }) = 1\]
anonymous
  • anonymous
So now we can out the new bounds in \[4 \int\limits_{0}^{\frac{ \pi }{ 2 }} \sqrt{r^2 - [(r) \sin \theta]^2 }\]
anonymous
  • anonymous
I meant to write put
anonymous
  • anonymous
If we distribute the square \[4 \int\limits_{0}^{\pi} \sqrt{r^2 - [r^2Sin^2]}\] now we can factor out the r^2
anonymous
  • anonymous
Okay
anonymous
  • anonymous
\[4 \int\limits_{0}^{\frac{ \pi }{ 2 }}\sqrt{r^2 (1 - \sin^2)}\] the square root of r squared is just r, so we can factor out an r \[4r \int\limits_{0}^{\frac{ \pi }{ 2 }} \sqrt{(1 - \sin \theta^2)}(r)\cos \theta d \theta\]
anonymous
  • anonymous
isn't 1-sin theta a trig identity
anonymous
  • anonymous
yes it = cos^2 theta
anonymous
  • anonymous
we can also pull the r (at the end) out to the front \[4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \sqrt{\cos^2 \theta} \cos \theta\]
anonymous
  • anonymous
The square root of cos^2 theta is just cos theta \[4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \cos \theta \cos \theta = 4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \cos^2 \theta\]
anonymous
  • anonymous
\[\cos^2 \theta = \frac{ \cos(2 \theta) }{ 2 } = \frac{ 1 }{ 2 } \cos(2 \theta)\] I forgot ot add the + 1 in the equation \[4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{ 2 } \cos (2 \theta) = \frac{ 1 }{ 2 }4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }}\cos(2 \theta)\] \[\int\limits \cos (2 \theta) + 1 = \frac{ 1 }{ 2 }\sin (2 \theta) + \theta\] \[2 r^2 [\frac{ 1 }{ 2 }\sin \theta + \theta]0 \rightarrow \frac{ \pi }{ 2 }\]
anonymous
  • anonymous
I made a mistake in the previous post, it should be \[2r^2 [\frac{ 1 }{ 2 } \sin (2 \theta + \theta)]\] with the same bounds now we need to plug in pi/2 and 0
anonymous
  • anonymous
\[2r^2 [[\frac{ 1 }{ 2 } \sin 2 \frac{ \pi }{ 2 } + \frac{ \pi }{ 2 }] - [\frac{ 1 }{ 2 }\sin 2(0) + 0]]\] 2(pi/2) = pi and sin (pi) = 0 \[2 r^2(0 + \frac{ \pi }{ 2 }- 0) = \pi r^2\] Because the 2's cancel
anonymous
  • anonymous
Yea a lot of writing

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