The area bounded by y=4-x2 and y=0,y=3 is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{16}{3}$
B
$\frac{22}{3}$
C
$\frac{26}{3}$
D
$\frac{28}{3}$

Solution:

The given curve is $y = 4 - x^{2}$ and the lines are $y = 0, y = 3$

$y = 0 \Rightarrow x = \pm 2$ and $x = 0 \Rightarrow y = 4$

The rough skectch is as below

The required area is area of shaded region

The shaded area is double the area of the region bounded by the curve $x = \sqrt{4 - y}$ , the lines $y = 0, y = 3$ and $y -$ axis.

Hence, the required area is $A = 2 \int_{0}^{3} \sqrt{4 - y} d y$

it gives

$\begin{array}{rcl} A & = & 2 \int_{0}^{3} \sqrt{4 - y} d y \\ = & 2 {[- \frac{{(4 - y)}^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{3} \\ = & - \frac{4}{3} (1 - 4^{\frac{3}{2}}) \\ = & \frac{4}{3} (8 - 1) \\ = & \frac{28}{3} \end{array}$

Therefore, the required area is $\frac{28}{3} square units$

The area bounded by $y = 4 - x^{2}$ and $y = 0, y = 3$ is

Solution:

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