Here is quiz 3.

**Problem 1** is Exercise 9.3.16 from the book. Here is a graph showing the cardioid and the circle. The easiest way to compute the required area is by subtracting the area of the circle from that of the cardioid. (Note that the circle is completely contained in the cardioid as, for each we have that This shows that in the portion of the circle in the first quadrant is within the portion of the cardiod there. The same holds in the fourth quadrant by symmetry. Of course, all of this also follows directly from the graph.)

The area of the cardioid is given by and the area of the circle is (note that as varies from to the whole circle is traveled once).

By symmetry, so the area we want is given by

Recall that so the integral is

Now note from the Cartesian graph of that so the area we want is just

**Problem 2** is Exercise 9.5.50 from the book. Recall that the standard polar equation of a line not going through the origin is given by

where is the distance from the line to the origin, and is the angle of the point in the line that realizes this distance, i.e., are the polar coordinates of the point on the line closest to the origin.

There are at least two ways we can proceed:

1) We can directly find the distance from the line to the origin. To do this, we recall that the line that goes through the origin and is perpendicular to where is given by The point in closest to the origin is the intersection of these two lines. In the case that interests us, this is the intersection of and so or or so We have found that the point is closest in to the origin. Its distance is Its angle is on the fourth quadrant and satisfies i.e.,

Putting this together, is the desired equation.

2) The other method we saw in lecture is based in the fact that

and consists in first writing the Cartesian equation of the line directly in polar coordinates, and then using the resulting expression to find and We have so or This gives us

We find a constant such that and are the cosine and the sine of some angle, precisely We then have

To find this we use that for any so must satisfy or so The value of such that and is This means that the equation we are looking for is

just as with the previous method.

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