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The area included between the parabolas y2 = 4x and x2 = 4y is (in square units)

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16/3

We have, \[x = \frac{y^2}{4} . . . . . \left( 1 \right)\]\[ x^2 = 4y . . . . . \left( 2 \right)\]Points of intersection of two parabola is given by,\[\left( \frac{y^2}{4} \right)^2 = 4y\]\[ \Rightarrow y^4 - 64y = 0\]\[ \Rightarrow y\left( y^3 - 64 \right) = 0\]\[ \Rightarrow y = 0, 4\]\[\Rightarrow x = 0, 4\]Therefore, the points of intersection are A(0, 0) and C(4, 4).Therefore, the area of the required region ABCD,\[= \int_0^4 \left( y_1 - y_2 \right) d x ..............\left(\text{where, }y_1 = 2\sqrt{x}\text{ and }y_2 = \frac{x^2}{4} \right)\]\[ = \int_0^4 \left( 2\sqrt{x} - \frac{x^2}{4} \right) d x\]\[ = \left[ 2 \times \frac{2 x^\frac{3}{2}}{3} - \frac{x^3}{12} \right]_0^4 \]\[ = \left( 2 \times \frac{2 \left( 4 \right)^\frac{3}{2}}{3} - \frac{\left( 4 \right)^3}{12} \right) - \left( 2 \times \frac{2 \left( 0 \right)^\frac{3}{2}}{3} - \frac{\left( 0 \right)^3}{12} \right)\]\[ = \left( \frac{32}{3} - \frac{16}{3} \right) - 0\]\[ = \frac{16}{3}\text{ square units }\]

shaalaa.comArea of the Region Bounded by a Curve and a Line Report Error Is there an error in this question or solution?Q 2Q 1Q 3Chapter 21: Areas of Bounded Regions - MCQ [Page 62]

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RD Sharma Mathematics [English] Class 12Chapter 21 Areas of Bounded RegionsMCQ | Q 2 | Page 62

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