KCET · Maths · Differentiation
The area of the region bounded by the curve \( y=x^{2} \) and the line \( y=16 \) is
- A \( \frac{32}{3} \) sq. units
- B \( \frac{256}{3} \) sq. units
- C \( \frac{64}{3} \) sq. units
- D \( \frac{128}{3} \) sq. units
Answer & Solution
Correct Answer
(B) \( \frac{256}{3} \) sq. units
Step-by-step Solution
Detailed explanation
Given curve, \(y=x^{2} \rightarrow(1)\)
and given line, \(y=16 \rightarrow(2)\)
Therefore, required area is given by 2 Area of region \(\mathrm{OAC}\)
\[
\begin{array}{l}
=2 \int_{0}^{16} \sqrt{y} d y \\
=2 \times \frac{2}{3} \times\left[y^{\frac{3}{2}}\right]_{0}^{16}=2 \times \frac{2}{3} \times(16)^{\frac{3}{2}} \\
=2 \times \frac{2}{3} \times(4)^{3}=\frac{256}{3}
\end{array}
\]
and given line, \(y=16 \rightarrow(2)\)
Therefore, required area is given by 2 Area of region \(\mathrm{OAC}\)
\[
\begin{array}{l}
=2 \int_{0}^{16} \sqrt{y} d y \\
=2 \times \frac{2}{3} \times\left[y^{\frac{3}{2}}\right]_{0}^{16}=2 \times \frac{2}{3} \times(16)^{\frac{3}{2}} \\
=2 \times \frac{2}{3} \times(4)^{3}=\frac{256}{3}
\end{array}
\]
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