-->

Vector Algebra

Question
CBSEENMA12032691

Find the area of the region lying in the first quadrant and bounded by x2 = y - 3, y = 4, y = 6 and the y-axis in the first quadrant.

Solution
The equation of curve is x2 = y - 3. which is an upward parabola whose vertex is (0,3).
Lines are y = 4,  y = 6
Required area = Area ABCD
                         equals space integral subscript 4 superscript 6 straight x space dy space equals space integral subscript 4 superscript 6 square root of straight y minus 3 end root space dy
                            equals space integral subscript 4 superscript 6 left parenthesis straight y minus 3 right parenthesis to the power of 1 half end exponent dy space equals space open square brackets fraction numerator left parenthesis straight y minus 3 right parenthesis 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 4 superscript 6
equals space 2 over 3 open square brackets left parenthesis straight y minus 3 right parenthesis to the power of 3 over 2 end exponent close square brackets subscript 4 superscript 6 space equals space 2 over 3 open square brackets left parenthesis 3 right parenthesis to the power of 3 over 2 end exponent minus left parenthesis 1 right parenthesis to the power of 3 over 2 end exponent close square brackets
space equals 2 over 3 left parenthesis 3 square root of 3 space minus space 1 right parenthesis space sq. space units.

Some More Questions From Vector Algebra Chapter