The correct answer is : 42
A = required area
\(=\int\limits^{\frac{3}{2}}_0\left[(3-y)-(\frac{2y^2}{3})\right]dy-\pi(\sqrt2)^2.\frac{1}{8}\)
\(⇒\left(3y-\frac{y^2}{2}-\frac{2}{9}y^3\right)\big|^{\frac{3}{2}}_{0}-\frac{\pi}{4}\)
\(⇒3.\frac{3}{2}-\frac{9}{8}-\frac{2}{9}.\frac{27}{8}-\frac{\pi}{4}\)
\(⇒\frac{36-9-6}{8}-\frac{\pi}{4}\)
\(=\frac{21}{8}-\frac{\pi}{4}\)
\(⇒4(\pi+4A)=4(\frac{21}{2})\)
\(=42\)
The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \)} and \( x = 2 \) is:
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:
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