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

[Steven G]

1/2*4 = 2. Simplified

 

[Kulla_9289]

It's 1/8.

 

[blamocur]

How do I remove the y, as the question only wants it in terms of a and ϕ?

Take a closer look at the post # 19 by @Dr.Peterson

 

[Kulla_9289]

I did. I got y in my new answer.

 

[Dr.Peterson]

I have thought about it and used a circle theorem to get to $(a)^2(π-2ϕ)+\frac{1}{2}(a^2)(4ϕ)-\frac{1}{2}(y^2)(ϕ)$.
Edit: I have simplified it to $πa^2-\frac{1}{2}y^2ϕ$. How do I remove the y, as the question only wants it in terms of a and ϕ?

I did. I got y in my new answer.

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?

 

[Kulla_9289]

$y=\sqrt{\frac{a^2π}{Φ}}$. And finally, got it to $πa^2-\frac{1}{2}(a^2π)$. No one deserves to lose a point here because of this.

 

[blamocur]

$y=\sqrt{\frac{a^2π}{Φ}}$. And finally, got it to $πa^2-\frac{1}{2}(a^2π)$. No one deserves to lose a point here because of this.

How did you get your expression for $y$ ?

 

[Kulla_9289]

I made it into an equation equated to 0; then brought half y^2 to the other side.

 

[Dr.Peterson]

I made it into an equation equated to 0; then brought half y^2 to the other side.

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?

 

[Kulla_9289]

This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.

 

[Otis]

1/2*4 = 2

It's 1/8.

1/2*4 = 2

1/(2*4) = 1/8

Order of Operations
$\;$

 

[Steven G]

1/2*4 = 2

1/(2*4) = 1/8

Order of Operations
$\;$

Was that you in my remedial arithmetic class last semester?

 

[blamocur]

This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.

Are you having difficulty with the triangle in the post #19? Have you tried to work it out? What have you tried to find $y$ ?

 

[Dr.Peterson]

This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.

Solve for $y$ in terms of $a$ and $\phi$. Then put the resulting expression into $a^2\pi-(...)^2\phi$.

But also, check whether what you want in your second term is really $\phi$.

 

[MaxMath]

From the sketches in #19 & #34 you should be able to express y in terms of a and $\phi$.

 

[Kulla_9289]

My approach:
1) $πr^2-\frac{1}{2}r^2θ$
2) $πa^2-\frac{1}{2}y^2ϕ$

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.

 

[Kulla_9289]

Here is the area of two shaded regions using your approach:

 

[Dr.Peterson]

Here is the area of two shaded regions using your approach:

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 $\phi$. I don't think you've even tried doing that yet, except perhaps when you said this:

$y=\sqrt{\frac{a^2π}{Φ}}$. And finally, got it to $πa^2-\frac{1}{2}(a^2π)$.

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.

 

[Kulla_9289]

Is the value of y = 2a cosϕ?

 

[Dr.Peterson]

Is the value of y = 2a cosϕ?

Yes. Now finish!