What is the area of a slice of a disk with angle theta?

[Physics197]

Homework Statement

Draw a circle of radius R with the center at the origin. Get the function that represents the top half of the circle. Let d be between 0 and R. State the definite integral that gives this area. Let P be the point were x=d intersects the curve. Let theta be the angle between the line and the horizontal, show that it equals cos^-1(d/R).

a) Give a formula for the area of a slice of a disk with angle theta.
b) Give a formula for the area of the triangle formed by P, the origin, and the point (d,0)
c) Now use this information to find a formula for: the integral(from d to R) of sqr(R^2-x^20

Homework Equations

R^2 = x^2 + y^2
Area of triangle =0.5bh

The Attempt at a Solution

What would a slice of a disk be? would the area be: Height*dx , where dx is the base?

Area of tri = 0.5 dRsin(theta)

what other formula do we need to find? the integral or another equation that is equal to the given one.

 

[n!kofeyn]

Did you answer all the questions in the paragraph preceding the (a)-(c)?

a) This is just the fraction of the total area of the disk that depends on the angle theta:
$$\text{Area of slice} = \frac{\theta}{2\pi}\pi r^2 = \frac{\theta r^2}{2}$$

b) You don't need to be finding differentials. Just explicitly give the formula.

c) Now can you find the formula?

 

[Physics197]

Yes I got all of those questions done.

How did you get that equation?

And I am not quite sure what you mean regarding the area of the triangle.Will the formula just simply be the area of the slice - the area of the triangle?

 

[n!kofeyn]

The area of a circle is $\pi r^2$. You can view this as $\frac{2\pi}{2\pi} \pi r^2$. The area of the sector of angle theta is proportional to the value of theta. So the area of the sector is just the fraction of the total area that depends on the fraction of $\frac{\theta}{2\pi}$. Search Google for alternate explanations.

What's the triangle's width? Its height?

 

[Physics197]

The area of a circle is $\pi r^2$. You can view this as $\frac{2\pi}{2\pi} \pi r^2$. The area of the sector of angle theta is proportional to the value of theta. So the area of the sector is just the fraction of the total area that depends on the fraction of $\frac{\theta}{2\pi}$. Search Google for alternate explanations.

What's the triangle's width? Its height?

Ohhh ok I get it. when they asked for the area of a slice of a disk, I didn't think of finding the area of the sector, I thought it was looking for really small rectangles with a width of dx, and the sum of these rectangles as the number of rectangles went to infinity equaled the area.