The area (in sq. units) of the region bounded by x^2=4y, y=1, y=9 and x=0 in the first quadrant is : Option 1: 104/3 Option 2: 36 Option 3: 112/3 Option 4: 124/3

Answer: 104/3. Solution: Integrate with respect to y using x = 2 y. Area = _1^9 2 y dy = 4/3[y^(3/2)]_1^9 = 4/3(27-1) = 104/3 sq. units.

Calculus

Application of Integrals

Medium

applied

The area (in sq. units) of the region bounded by $x^2=4y, y=1, y=9$ and $x=0$ in the first quadrant is :

$\dfrac{104}{3}$

$\dfrac{112}{3}$

$\dfrac{124}{3}$

✅ Correct Option: 1

Integrate with respect to $y$ using $x = 2\sqrt y$ .

Area $= \int_1^9 2\sqrt y\,dy = \dfrac{4}{3}\left[y^{3/2}\right]_1^9 = \dfrac{4}{3}(27-1) = \dfrac{104}{3}$ sq. units.

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