Key Concepts and Formulas

• The area between two curves $x = f(y)$ and $x = g(y)$ from $y=a$ to $y=b$ is given by $A = \int_a^b |f(y) - g(y)| dy$. We determine which function is the "rightmost" curve ($x_{right}$) and the "leftmost" curve ($x_{left}$) to evaluate the integral as $A = \int_a^b (x_{right} - x_{left}) dy$.
• $\int \frac{1}{1+y^2} dy = \arctan(y) + C$
• If $f(y)$ is an even function, i.e., $f(-y) = f(y)$, then $\int_{-a}^{a} f(y) dy = 2 \int_{0}^{a} f(y) dy$.

Step-by-Step Solution

Step 1: Express the Curves as Functions of y

We are given the equations $x(1+y^2)=1$ and $y^2=2x$. To integrate with respect to $y$, we rewrite both equations in the form $x = f(y)$.

• Equation 1: $x(1+y^2) = 1 \implies x = \frac{1}{1+y^2}$
• Equation 2: $y^2 = 2x \implies x = \frac{y^2}{2}$

Step 2: Find the Intersection Points

To find the points of intersection, we set the two expressions for $x$ equal to each other: $\frac{1}{1+y^2} = \frac{y^2}{2}$ Cross-multiplying gives: $2 = y^2(1+y^2)$ $2 = y^2 + y^4$ Rearranging into a quartic equation: $y^4 + y^2 - 2 = 0$ Let $u = y^2$. Then the equation becomes: $u^2 + u - 2 = 0$ Factoring the quadratic equation: $(u+2)(u-1) = 0$ So, $u = -2$ or $u = 1$. Since $u = y^2$, we have $y^2 = -2$ or $y^2 = 1$. $y^2 = -2$ has no real solutions. $y^2 = 1$ gives $y = \pm 1$.

Now, we find the corresponding $x$ values. Using $x = \frac{y^2}{2}$:

• When $y = 1$, $x = \frac{(1)^2}{2} = \frac{1}{2}$.
• When $y = -1$, $x = \frac{(-1)^2}{2} = \frac{1}{2}$. Thus, the points of intersection are $(\frac{1}{2}, 1)$ and $(\frac{1}{2}, -1)$. The limits of integration with respect to $y$ will be from $-1$ to $1$.

Step 3: Determine the Rightmost and Leftmost Curves

We need to determine which curve is to the right and which is to the left in the region of interest. Let's consider $y=0$, which is between -1 and 1.

• For $x = \frac{1}{1+y^2}$, when $y=0$, $x = \frac{1}{1+0^2} = 1$.
• For $x = \frac{y^2}{2}$, when $y=0$, $x = \frac{0^2}{2} = 0$. Since $1 > 0$, the curve $x = \frac{1}{1+y^2}$ is to the right of the curve $x = \frac{y^2}{2}$ in the region between $y=-1$ and $y=1$. Therefore, $x_{right} = \frac{1}{1+y^2}$ and $x_{left} = \frac{y^2}{2}$.

Step 4: Set Up the Definite Integral

The area $A$ is given by: $A = \int_{-1}^{1} (x_{right} - x_{left}) dy = \int_{-1}^{1} \left(\frac{1}{1+y^2} - \frac{y^2}{2}\right) dy$ Since the integrand is an even function and the limits are symmetric, we can write: $A = 2 \int_{0}^{1} \left(\frac{1}{1+y^2} - \frac{y^2}{2}\right) dy$

Step 5: Evaluate the Integral

Now, we evaluate the integral: $A = 2 \left[ \int_{0}^{1} \frac{1}{1+y^2} dy - \int_{0}^{1} \frac{y^2}{2} dy \right]$ $A = 2 \left[ \arctan(y) \Big|_0^1 - \frac{y^3}{6} \Big|_0^1 \right]$ $A = 2 \left[ (\arctan(1) - \arctan(0)) - \left(\frac{1^3}{6} - \frac{0^3}{6}\right) \right]$ $A = 2 \left[ \left(\frac{\pi}{4} - 0\right) - \left(\frac{1}{6} - 0\right) \right]$ $A = 2 \left(\frac{\pi}{4} - \frac{1}{6}\right)$ $A = \frac{\pi}{2} - \frac{1}{3}$

Step 6: Conclusion

The area of the region bounded by the curves $x(1+y^2)=1$ and $y^2=2x$ is $\frac{\pi}{2} - \frac{1}{3}$.

Common Mistakes & Tips

• Failing to correctly identify the rightmost and leftmost curves can lead to a negative area or an incorrect result. Always test a point within the interval or sketch the curves.
• Incorrectly evaluating the definite integral, especially the arctangent function, is a common error. Remember that $\arctan(1) = \frac{\pi}{4}$ and $\arctan(0) = 0$.
• Forgetting to multiply by 2 when using symmetry can lead to an answer that's half the correct value.

Summary

We expressed both curves as functions of $y$, found their intersection points, determined which curve was to the right and which was to the left, set up the definite integral, and evaluated it to find the area. By utilizing the symmetry of the region, we simplified the calculation. The area of the region is $\frac{\pi}{2} - \frac{1}{3}$.

Final Answer

The final answer is $\boxed{\frac{\pi}{2} - \frac{1}{3}}$, which corresponds to option (B).