The area of the region bounded by the curve
\( x = 4 - y^{2} \)
and the y-axis is:
Show Hint
- $32/3$
- $16/3$
- $8/3$
- $64/3$
The Correct Option is A
Solution and Explanation
Area $= \int_{c}^{d} x dy$.
Step 2: Analysis
The curve intersects the y-axis where $x = 0$, so $4 - y^2 = 0 \Rightarrow y = \pm 2$.
Area $= \int_{-2}^{2} (4 - y^2) dy = [4y - y^3/3]_{-2}^{2}$.
Step 3: Conclusion
$= (8 - 8/3) - (-8 + 8/3) = 16 - 16/3 = 32/3$.
Final Answer: (A)
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