Question

The area of the region bounded by the curve \( x = 4 - y^{2} \) and the y-axis is:

• A. $32/3$
• B. $16/3$
• C. $8/3$
• D. $64/3$

Hint:

This parabola is symmetric about the x-axis, so you can calculate $\int_{0}^{2} x dy$ and double the result.

Answer 1

The Correct Option is A

Step 1: Understand the Concept
To find the area of a region bounded by a curve and the y-axis, we use integration with respect to $y$. If the curve is defined as $x = f(y)$, the area between the horizontal lines $y = c$ and $y = d$ is given by the definite integral:
Area = $\int_{c}^{d} x \, dy$
In this case, the curve is $x = 4 - y^2$, which is a parabola opening to the left.

Step 2: Analysis & Limits of Integration
First, we need to find the points where the curve intersects the y-axis (where the boundaries of our area lie). On the y-axis, the value of $x$ is always 0. Setting the equation to zero, we get:
$4 - y^2 = 0 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.
Therefore, our limits of integration are from $y = -2$ to $y = 2$.

Now, we set up the integral:
Area = $\int_{-2}^{2} (4 - y^2) \, dy$
Integrating the terms individually:
Area = $[4y - \frac{y^3}{3}]_{-2}^{2}$

Step 3: Calculation & Conclusion
We now apply the Fundamental Theorem of Calculus by substituting the upper limit and subtracting the substitution of the lower limit:
Upper limit ($y=2$): $4(2) - \frac{2^3}{3} = 8 - \frac{8}{3}$
Lower limit ($y=-2$): $4(-2) - \frac{(-2)^3}{3} = -8 - (-\frac{8}{3}) = -8 + \frac{8}{3}$

Subtracting them:
Area = $(8 - \frac{8}{3}) - (-8 + \frac{8}{3})$
Area = $8 - \frac{8}{3} + 8 - \frac{8}{3} = 16 - \frac{16}{3}$
To simplify, convert 16 to a fraction: $\frac{48}{3} - \frac{16}{3} = \frac{32}{3}$.
The area of the bounded region is $\frac{32}{3}$ square units.

Final Answer: (A)