The ﬁgure on the right shows the ellipse $\frac{(x-19)^2}{19}+\frac{(x-98)^2}{98}=1998$.
Let $R_1,R_2,R_3,$ and $R_4$ denote those areas within the ellipse that are in the ﬁrst, second, third, and fourth quadrants, respectively. Determine the value of $R_1-R_2+R_3-R_4$.

[asy]
defaultpen(linewidth(0.7));
pair c=(19,98);
real dist = 30;
real a = sqrt(1998*19),b=sqrt(1998*98);
xaxis("x",c.x-a-dist,c.x+a+3*dist,EndArrow);
yaxis("y",c.y-b-dist*2,c.y+b+3*dist,EndArrow);
draw(ellipse(c,a,b));
label("$R_1$",(100,200));
label("$R_2$",(-80,200));
label("$R_3$",(-60,-150));
label("$R_4$",(70,-150));[/asy]
1. **Correct the Ellipse Equation:**
   The given equation of the ellipse is:
   \[
   \frac{(x-19)^2}{19} + \frac{(x-98)^2}{98} = 1998
   \]
   This appears to be a typo. The correct equation should be:
   \[
   \frac{(x-19)^2}{19} + \frac{(y-98)^2}{98} = 1998
   \]

2. **Transform the Equation:**
   We transform the given equation to the standard form of an ellipse:
   \[
   \frac{(x-19)^2}{19 \cdot 1998} + \frac{(y-98)^2}{98 \cdot 1998} = 1
   \]
   This simplifies to:
   \[
   \frac{(x-19)^2}{37962} + \frac{(y-98)^2}{195804} = 1
   \]

3. **Calculate the Area of the Ellipse:**
   The area \(A\) of an ellipse is given by:
   \[
   A = \pi \cdot a \cdot b
   \]
   where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. Here:
   \[
   a = \sqrt{195804} = 198.95 \quad \text{and} \quad b = \sqrt{37962} = 194.83
   \]
   Therefore, the area of the ellipse is:
   \[
   A = \pi \cdot 198.95 \cdot 194.83 = 122,000\pi
   \]

4. **Determine \(R_1 - R_2 + R_3 - R_4\):**
   The areas \(R_1, R_2, R_3,\) and \(R_4\) are the areas of the ellipse in the first, second, third, and fourth quadrants, respectively. By symmetry, each quadrant has an equal area:
   \[
   R_1 = R_2 = R_3 = R_4 = \frac{A}{4} = \frac{122,000\pi}{4} = 30,500\pi
   \]
   Therefore:
   \[
   R_1 - R_2 + R_3 - R_4 = 30,500\pi - 30,500\pi + 30,500\pi - 30,500\pi = 0
   \]

The final answer is \(\boxed{0}\)