Question

Prove the circumference of a circle is 2πr
@LeibyStrauss

Answer 1

The formula for a cirle is x^2 + y^2 = r^2

To solve for y: - x^2 from both sides

Answer 2

\[y^2 = r^2 - x^2\]

take the square root of both sides

\[y = \sqrt{r^2 - x^2}\]

Answer 3

integrate

Answer 4

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)

Answer 5

correct

Answer 6

okay i see

Answer 7

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

\[4 \int\limits \sqrt{r^2 - [(r)\sin \theta]^2} (r)\cos \theta\]

Answer 8

\[x = rsin\]

Answer 9

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

Answer 10

\[r = (r) \sin \theta\]
divide both sides by r
\[1 = \sin \theta\]
\[\theta = \frac{ \pi }{ 2 }\]
because
\[\sin (\frac{ \pi }{ 2 }) = 1\]

Answer 11

So now we can out the new bounds in

\[4 \int\limits_{0}^{\frac{ \pi }{ 2 }} \sqrt{r^2 - [(r) \sin \theta]^2 }\]

Answer 12

I meant to write put

Answer 13

If we distribute the square

\[4 \int\limits_{0}^{\pi} \sqrt{r^2 - [r^2Sin^2]}\]

now we can factor out the r^2

Answer 14

Okay

Answer 15

\[4 \int\limits_{0}^{\frac{ \pi }{ 2 }}\sqrt{r^2 (1 - \sin^2)}\]

the square root of r squared is just r, so we can factor out an r

\[4r \int\limits_{0}^{\frac{ \pi }{ 2 }} \sqrt{(1 - \sin \theta^2)}(r)\cos \theta d \theta\]

Answer 16

isn't 1-sin theta a trig identity

Answer 17

yes it = cos^2 theta

Answer 18

we can also pull the r (at the end) out to the front

\[4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \sqrt{\cos^2 \theta} \cos \theta\]

Answer 19

The square root of cos^2 theta is just cos theta

\[4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \cos \theta \cos \theta = 4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \cos^2 \theta\]

Answer 20

\[\cos^2 \theta = \frac{ \cos(2 \theta) }{ 2 } = \frac{ 1 }{ 2 } \cos(2 \theta)\]

I forgot ot add the + 1 in the equation

\[4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{ 2 } \cos (2 \theta) = \frac{ 1 }{ 2 }4r^2 \int\limits_{0}^{\frac{ \pi }{ 2 }}\cos(2 \theta)\]

\[\int\limits \cos (2 \theta) + 1 = \frac{ 1 }{ 2 }\sin (2 \theta) + \theta\]

\[2 r^2 [\frac{ 1 }{ 2 }\sin \theta + \theta]0 \rightarrow \frac{ \pi }{ 2 }\]

Answer 21

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

Answer 22

\[2r^2 [[\frac{ 1 }{ 2 } \sin 2 \frac{ \pi }{ 2 } + \frac{ \pi }{ 2 }] - [\frac{ 1 }{ 2 }\sin 2(0) + 0]]\]

2(pi/2) = pi and sin (pi) = 0

\[2 r^2(0 + \frac{ \pi }{ 2 }- 0) = \pi r^2\]

Because the 2's cancel

Answer 23

Yea a lot of writing