The area of the region bounded by the parabol | vedclass

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(B) The equation of the parabola is $x^{2}=16y$,which gives $x=4\sqrt{y}$ for the first quadrant.The area $A$ is bounded by the curve $x=4\sqrt{y}$,th

Vedclass · Accepted Answer

$\frac{56}{3} 	ext{ sq. units}$

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(B) The equation of the parabola is $x^{2}=16y$,which gives $x=4\sqrt{y}$ for the first quadrant.The area $A$ is bounded by the curve $x=4\sqrt{y}$,the $Y$-axis,and the lines $y=1$ and $y=4$.$A = \int_{1}^{4} x \, dy = \int_{1}^{4} 4\sqrt{y} \, dy$$A = 4 \int_{1}^{4} y^{1/2} \, dy$$A = 4 \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4}$$A = 4 	imes \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$$A = \frac{8}{3} [4^{3/2} - 1^{3/2}]$$A = \frac{8}{3} [8 - 1]$$A = \frac{8}{3} 	imes 7 = \frac{56}{3} 	ext{ sq. units}$

The area of the region bounded by the parabola $x^{2}=16y$,the lines $y=1$,$y=4$,and the $Y$-axis in the first quadrant is:

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