Question

Find the area of the region bounded by x2 = 16y, y - axis and the lines y = 2 and y = 4.​

Answer 1

SOLUTION

TO DETERMINE

The area of the region bounded by x² = 16y , y - axis and the lines y = 2 and y = 4.

EVALUATION

Here the given equation of the curve is

x² = 16y

Now in the graph ABCDA is the region bounded by x² = 16y , y - axis and the lines y = 2 and y = 4.

Hence the required area

$\displaystyle \sf{= \int\limits_{2}^{4} x \, dy }$

$\displaystyle \sf{= \int\limits_{2}^{4} \sqrt{16y} \, dy }$

$\displaystyle \sf{=4 \int\limits_{2}^{4} {y}^{ \frac{1}{2} } \, dy }$

$\displaystyle \sf{=4 \times \frac{ {y}^{ \frac{3}{2} } }{ \frac{3}{2} } \bigg|_{2}^{4} }\: \: \: Sq. Unit$

$\displaystyle \sf{=4 \times \frac{2}{3} \times {y}^{ \frac{3}{2} } \bigg|_{2}^{4} } \: \: \: \: Sq. Unit$

$\displaystyle \sf{= \frac{8}{3} \times \bigg( {4}^{ \frac{3}{2} } - {2}^{ \frac{3}{2} } \bigg) }\: \: \: Sq. Unit$

$\displaystyle \sf{= \frac{8}{3} \times \bigg( 8 - 2 \sqrt{2} \bigg) }\: \: \: Sq. Unit$

$\displaystyle \sf{= \frac{8}{3} \bigg( 8 - 2 \sqrt{2} \bigg) } \: \: \: Sq. Unit$

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