Is my answer wrong? I got (1/2)a^2(2π−sin(2Ø)−2cos(Ø))

Kulla_9289 said:
I have thought about it and used a circle theorem to get to [imath](a)^2(π-2ϕ)+\frac{1}{2}(a^2)(4ϕ)-\frac{1}{2}(y^2)(ϕ)[/imath].
Edit: I have simplified it to [imath]πa^2-\frac{1}{2}y^2ϕ[/imath]. How do I remove the y, as the question only wants it in terms of a and ϕ?
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Kulla_9289 said:
I did. I got y in my new answer.
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But you don't want y in the answer.

You've already been told that you can express y in terms of a and phi; and I showed you a picture to help with that. Do that, substitute for y, and you will not have y in the answer; everything will be in terms of a and phi. Do you see this?
 
[imath]y=\sqrt{\frac{a^2π}{Φ}}[/imath]. And finally, got it to [imath]πa^2-\frac{1}{2}(a^2π)[/imath]. No one deserves to lose a point here because of this.
 
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Kulla_9289 said:
[imath]y=\sqrt{\frac{a^2π}{Φ}}[/imath]. And finally, got it to [imath]πa^2-\frac{1}{2}(a^2π)[/imath]. No one deserves to lose a point here because of this.
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How did you get your expression for [imath]y[/imath] ?
 
I made it into an equation equated to 0; then brought half y^2 to the other side.
 
Kulla_9289 said:
I made it into an equation equated to 0; then brought half y^2 to the other side.
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Please show the details of your work, so we can see where you are thinking incorrectly. What was that equation, and why did you think it was valid?
 
This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.
 
Kulla_9289 said:
This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.
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Are you having difficulty with the triangle in the post #19? Have you tried to work it out? What have you tried to find [imath]y[/imath] ?
 
Kulla_9289 said:
This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.
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Solve for [imath]y[/imath] in terms of [imath]a[/imath] and [imath]\phi[/imath]. Then put the resulting expression into [imath]a^2\pi-(...)^2\phi[/imath].

But also, check whether what you want in your second term is really [imath]\phi[/imath].
 
From the sketches in #19 & #34 you should be able to express y in terms of a and [imath]\phi[/imath].
 
My approach:
1) [imath]πr^2-\frac{1}{2}r^2θ[/imath]
2) [imath]πa^2-\frac{1}{2}y^2ϕ[/imath]

I have also used your triangle approach to get the same answer. There will be 4 of these triangles you have sketched out. The area of all 4 triangles gets cancelled out later on.
 
Here is the area of two shaded regions using your approach:
 

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Kulla_9289 said:
Here is the area of two shaded regions using your approach:
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You are misunderstanding what "my approach" is.

I am not (primarily) saying that your formula with y is wrong, or should be obtained a different way; I am telling you to solve that little right triangle for the value of y, given known values of a and [imath]\phi[/imath]. I don't think you've even tried doing that yet, except perhaps when you said this:
Kulla_9289 said:
[imath]y=\sqrt{\frac{a^2π}{Φ}}[/imath]. And finally, got it to [imath]πa^2-\frac{1}{2}(a^2π)[/imath].
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That formula for y is wrong; but you never showed us the steps you took to get it, so we could help you correct it.
 
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