A) 16 sq unit
B) 32 sq unit
C) \[\frac{32}{3}\]sq unit
D) \[\frac{16}{3}\]sq unit
Correct Answer: C
Solution :
Given, curve \[x=4-{{y}^{2}}\] and y-axis. The required area is\[ABCOA\] \[2\times \]area of\[ABOA\] \[=2\times \int_{0}^{4}{\sqrt{4-x}\,\,dx}\] \[=2\left[ -\frac{{{(4-x)}^{3/2}}}{3/2} \right]_{0}^{4}\] \[=2\left[ -\frac{2}{3}\times 0+\frac{2}{3}{{(4)}^{3/2}} \right]\] \[=\frac{4}{3}\times 8\] \[=\frac{32}{3}\text{sq}\,\,\text{unit}\]You need to login to perform this action.
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