Find the area of the region bounded by x 2 = 16 y, y = 1, y = 4 and the y-axis in the first quadrant. - Wired Faculty

Vector Algebra

Question

Find the area of the region bounded by x² = 16 y, y = 1, y = 4 and the y-axis in the first quadrant.

Solution

The equation of curve is
$straight x squared space equals space 16 space straight y$
Required area = $integral subscript 1 superscript 4 straight x space dy$
$equals space integral subscript 1 superscript 4 4 square root of straight y dy space equals space 4 integral subscript 1 superscript 4 straight y to the power of 1 half end exponent dy space equals space 4 open square brackets fraction numerator straight y to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 1 superscript 4 equals space 8 over 3 open square brackets straight y to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 4 space equals space 8 over 3 open square brackets left parenthesis 4 right parenthesis to the power of 3 over 2 end exponent minus 1 close square brackets space equals space 8 over 3 left square bracket 8 minus 1 right square bracket space equals 8 over 3 cross times 7 space equals space 56 over 3 space sq. space units.$

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Vector Algebra

Find the area of the region bounded by x² = 16 y, y = 1, y = 4 and the y-axis in the first quadrant.

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Vector Algebra

Find the area of the region bounded by x2 = 16 y, y = 1, y = 4 and the y-axis in the first quadrant.

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the area of the region bounded by x² = 16 y, y = 1, y = 4 and the y-axis in the first quadrant.