W15 - HW Solutions

(a)

(b)

The quadratic formula provides all complex roots, using that :

First convert this polar curve to a parametric curve using and :

Then use . Differentiate:

Therefore:

(a) Find the angle of the line from the origin to the point of intersection of the two curves (in Quadrant I):

Compute the area below this line, inside the larger circle, and above the -axis:

(This circular sector is also just of the whole disk area, which is .)

Compute the area above the line and inside the smaller circle:

Combined area in green above the -axis is . Double this for the total green area:

(b) Notice that green and yellow combine to give the area of the smaller circle. The area of the smaller circle is .

Therefore, the yellow region has area:

Note: It is also reasonable to find the yellow region first, using this formula:

Find the intersection between the line and the curve, in Quadrant I:

Of course , so we have and (in Quadrant I) therefore . To get the answer, we double the area from this angle up to the vertical :

Solve for consecutive (in ) solutions to to get the starting and ending for a single loop:

Integrate:

Solve for consecutive (in ) solutions to to get the starting and ending for a single loop:

The interval corresponds to the inner loop. To see this, draw a graph of the limaçon:

(a)

(b)

The correct interpretation is and and therefore .

It would not be correct to write .

Each instance of the symbol “” involves making a choice of root. There are possible choices.

(a)

(b)

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