Question

Which of the following expressions will give the area of the region bounded by the curve $y = x^2$ and the line $y = 16$?

• A. $\int_{0}^{4} 2x \, dx$
• B. $2 \int_{0}^{4} 2x \, dx$
• C. $\int_{0}^{16} y \, dy$
• D. $2 \int_{0}^{16} \sqrt{y} \, dy$

Hint:

If integrated along the $x$-axis, the area would be $\int_{-4}^{4} (16 - x^2) \, dx = 2 \int_{0}^{4} (16 - x^2) \, dx$. Always verify whether the options integrate with respect to $x$ or $y$ to choose the matching formulation.

Answer 1

The Correct Option is D

Concept: The area of a region bounded by a curve can be integrated either along the $x$-axis or along the $y$-axis.
• Integrating along the $y$-axis: $\text{Area} = \int_{c}^{d} x \, dy$, where $x$ is expressed as a function of $y$.
• Since the parabola $y = x^2$ is symmetric about the $y$-axis, the total area is twice the area on the positive side of the $y$-axis.

Step 1: Understand the boundary and symmetry of the region.
The given curve is a vertical upward-opening parabola $y = x^2$. The line bounding it from above is $y = 16$. The points of intersection are found by equating the two expressions: \[ x^2 = 16 \quad \Rightarrow \quad x = \pm 4 \] Thus, the region extends along the $x$-axis from $x = -4$ to $x = 4$, and along the $y$-axis from $y = 0$ to $y = 16$. Because a parabola is perfectly symmetric about the $y$-axis, we can compute the area of the right-hand side (where $x \ge 0$) and multiply it by 2.

Step 2: Express the area formula by integrating with respect to $y$.
If we choose horizontal elements to integrate along the $y$-axis from the base $y = 0$ to the top line $y = 16$: From $y = x^2$, since $x \ge 0$ in the first quadrant, we can isolate $x$: \[ x = \sqrt{y} \] The bounded region spans from $y=0$ to $y=16$. Due to symmetry, the total area $A$ is: \[ A = 2 \times \int_{0}^{16} x \, dy \] Substituting $x = \sqrt{y}$ into the equation gives: \[ A = 2 \int_{0}^{16} \sqrt{y} \, dy \] This matches perfectly with choice (D).