what is the area of region R?

Question

the region R is bounded by lines y=(1/2)x and y=(1/3)x and the curve given by y^2=10x-x^2. Line y=(1/2)x intersects the curve at point A, and line y=(1/3)x intersects the curve at point B. I know it is easy to understand with a figure, but i don't know how to post it on line from the paper. sorry!
(a) show the relation y^2=10x-x^2 can be expressed in polar coordinates as r=10cosx?
(b) Use the polar equation in part a to write down an integral expression with respect to the polar angle x to represent the area of region R.

Answer 1

EDIT FOR IRONDUKE'S ANSWER BELOW: Your area integral is not set up correctly.


BEGIN MY ANSWER:

Part a) is easy. The key conversions between polar and rectangualr are

x = r cos Q (I use Q for "theta")
y = r sin Q
r^2 = x^2 + y^2.

You have y^2 = 10x - x^2. It would be convenient to have the x^2 on the left, so move it to get

x^2 + y^2 = 10x.

Why is it convenient? Because now the left side exactly matches the conversion to r^2. Also, the x on the right side can be converted to r cos Q (you wrote r cos x in your question, but it's not x, it's theta).

Make these two conversions in your equation and you get

r^2 = 10 r cos Q.

Divide by r to get your desired result.

b) Trickier. The area of r(Q) (r as a function of theta) from Q = Q1 to Q = Q2 is given by 1/2 times the integral of r^2 dQ, from Q1 to Q2. Pretend the S below is an integral sign, with lower limit Q1 and upper limit Q2:

(1/2) S (r^2) dQ

The line y = 1/3 x, when converted to polar form, would be

r sin Q = 1/3 r cos Q.

Divide by r cos Q. The r's cancel, and sin/cos = tan. So you get

tan Q = 1/3.

Thus, Q = arctan(1/3), which is about .32 radians, or 18 degrees.

For the other line, you would get Q = arctan(1/2), which is about .46 radians, or 26 degrees.

These are your limits of integration. The lower one will be .32, and the upper .46. If you want to be exact, the lower limit will be arctan(1/3) and the upper will be arctan (1/2), i.e.

Q1 = arctan (1/3) and Q2 = arctan (1/2).

Now, since your equation from part a) is r = 10 cos Q, you integral will be

1/2 S (10 cosQ)^2 dQ, or
1/2 S 100 cos^2 Q dQ, or
50 S cos^2 Q dQ.

(Remember: the S is an integral sign, and the limits of integration are Q1 and Q2.

That should suffice for part b). I find it funny that the problem doesn't ask you to find the area. It's not a bad integral to finish, and even having funky limits of integration like Q1 = arctan (1/3) and Q2 = arctan (1/2) are managable.


Hope this helps.


EDIT: I found the area, since I took it this far. It's exactly

25 arctan (1/2) - 25 arctan (1/3) + 5/2

It took about 3 minutes, but I would expect a calculus student to take longer (I teach this stuff)).

Answer 2

y^2=10x-x^2 --> (x-5)^2+y^2 =25, so this is a circle with center at (5,0) and radius =5.

The line y =x/2 intersects this circle at x^2/4 =10x-x^2
x^2 =40x -4x^2, so 5x(x-8) = 0 --> x=8 and y = 4.

The line y=x/3 intersects the circle at x^2/9 =90x -x^2
x^2=90x - 9x^2 so 10x(x-9) --> x = 9 and y = 3
So Point A =(8,4) and Point B = (9,3).

The x-intercepts of the circle are (0,0) and (10,0).
The The triangles formed by drawing lines from A and B to each x-intercept are right triangles because the angles A and B are inscribed in a semicircle.

Thus r, the line from the origin to the circle is 10 cos x , where x is the angle that the line drawn from the origin to the point A, or any other point on the circle makes with the x-axis.

The angle x is arctan (1/3) for the line y= x/3 and arctan (1/2( for the line y=x/2

So the integral would be:
integral from arctan(1/3) to arctan(1/2)[.5*10cosxdx