The area of the region bounded by x^2=4y,y=1, | vedclass

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(B) The region is bounded by the parabola $x^2=4y$,the lines $y=1$ and $y=4$,and the y-axis in the first quadrant.From $x^2=4y$,we have $x = \sqrt{4y}

Vedclass · Accepted Answer

$\frac{28}{3}$

Vedclass · Answer

(B) The region is bounded by the parabola $x^2=4y$,the lines $y=1$ and $y=4$,and the y-axis in the first quadrant.From $x^2=4y$,we have $x = \sqrt{4y} = 2\sqrt{y}$ (since $x > 0$ in the first quadrant).The required area $A$ is given by the integral with respect to $y$:$A = \int_{1}^{4} x \, dy = \int_{1}^{4} 2\sqrt{y} \, dy$$A = 2 \int_{1}^{4} y^{1/2} \, dy$$A = 2 \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4}$$A = 2 	imes \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$$A = \frac{4}{3} (4^{3/2} - 1^{3/2})$$A = \frac{4}{3} (8 - 1)$$A = \frac{4}{3} (7) = \frac{28}{3}$ square units.

The area of the region bounded by $x^2=4y$,$y=1$,$y=4$ and the y-axis lying in the first quadrant is $ . . . . . . $ square units.

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