Finding equation for chord in terms of radius given angle

[burt]

In this problem, I am trying to find the volume of the solid gotten by rotating the shaded area around the x-axis. The equation of a circle is [MATH]x^2+y^2=r^2[/MATH]. If I am integrating using the shell method, I know the height and radius that I need (height = [MATH]\sqrt{r^2-y^2}[/MATH] and radius = y). My upper limit of integration is r. My lower limit of integration is c. But, how do I figure out how to put c in terms of r so I can simplify?

 

[Dr.Peterson]

I don't understand. Isn't c just a given constant? Why would you put it in terms of r?

Maybe you need to quote the entire problem as given to you.

EDIT: Ah! I see you are given the angle theta, and have to determine c from that (not from r alone).

Look at the right triangle with the shown radius as hypotenuse. What is the opposite side?

 

[pka]

View attachment 15602In this problem, I am trying to find the volume of the solid gotten by rotating the shaded area around the x-axis. The equation of a circle is [MATH]x^2+y^2=r^2[/MATH]. If I am integrating using the shell method, I know the height and radius that I need (height = [MATH]\sqrt{r^2-y^2}[/MATH] and radius = y). My upper limit of integration is r. My lower limit of integration is c. But, how do I figure out how to put c in terms of r so I can simplify?

The polar equation of that semicircle is \(\displaystyle R\cos(\theta),R\sin(\theta),~0\le\theta\le \pi\).
Note that \(\displaystyle y=c\) intersects the semicircle at \(\displaystyle \left(\frac{R\sqrt3}{2},\frac{R}{2}\right)\) i.e. \(\displaystyle \theta=\frac{\pi}{6}\)
Can you continue?

 

[burt]

I don't understand. Isn't c just a given constant? Why would you put it in terms of r?

Maybe you need to quote the entire problem as given to you.

EDIT: Ah! I see you are given the angle theta, and have to determine c from that (not from r alone).

Look at the right triangle with the shown radius as hypotenuse. What is the opposite side?

Oh, I'm starting to understand... C is the opposite side. Does that mean I can use the sine function?

 

[Dr.Peterson]

Yes. And the sine of 30° is particularly simple.