KCET · Maths · Area Under Curves
The area bounded by the curve \(x=4-y^{2}\) and the \(Y\)-axis is
- A \(16 \mathrm{sq}\) units
- B 32 sq units
- C \(\frac{32}{3}\) sq units
- D \(\frac{16}{3}\) sq units
Answer & Solution
Correct Answer
(C) \(\frac{32}{3}\) sq units
Step-by-step Solution
Detailed explanation
Given curve \(x=4-y^{2}\) and \(Y\)-axis.
The formed area is \(A B C O A\)
\[
\begin{aligned}
&=2 \times \text { area of } A B O A \\
&=2 \times \int_{0}^{4} \sqrt{4-x} d x \\
&=2\left[-\frac{(4-x)^{3 / 2}}{3 / 2}\right]_{0}^{4} \\
&=2\left[-\frac{2}{3} \times 0+\frac{2}{3}(4)^{3 / 2}\right] \\
&=\frac{4}{3} \times 8=\frac{32}{3} \text { sq units }
\end{aligned}
\]
The formed area is \(A B C O A\)
\[
\begin{aligned}
&=2 \times \text { area of } A B O A \\
&=2 \times \int_{0}^{4} \sqrt{4-x} d x \\
&=2\left[-\frac{(4-x)^{3 / 2}}{3 / 2}\right]_{0}^{4} \\
&=2\left[-\frac{2}{3} \times 0+\frac{2}{3}(4)^{3 / 2}\right] \\
&=\frac{4}{3} \times 8=\frac{32}{3} \text { sq units }
\end{aligned}
\]
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