Question

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

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

Hint:

When the region is symmetric about the y-axis, the width becomes \(2x\). Express the curve as \(x\) in terms of \(y\) and integrate with respect to \(y\).

Answer 1

The Correct Option is D

Step 1: Understand the region.
The curve is \[ y=x^2 \] and the line is \[ y=16 \] To find the intersection:
\[ x^2=16 \] \[ x=\pm4 \] Thus the region extends from \(x=-4\) to \(x=4\).
Step 2: Express $x$ in terms of $y$.
From \[ y=x^2 \] \[ x=\pm\sqrt{y} \] So the horizontal width of the region is
\[ \sqrt{y}-(-\sqrt{y}) \] \[ =2\sqrt{y} \] Step 3: Area using integration with respect to $y$.
Area formula becomes
\[ \int_{0}^{16} 2\sqrt{y}\,dy \] This can also be written as
\[ 2\int_{0}^{16}\sqrt{y}\,dy \] Step 4: Conclusion.
Therefore the correct expression representing the area is
\[ 2\int_{0}^{16}\sqrt{y}\,dy \] Final Answer: $\boxed{2\int_{0}^{16}\sqrt{y}\,dy}$