Answer to Question #114636 in Calculus for ANJU JAYACHANDRAN

Question #114636
Find the area between the curves y^2=ax and ay^2=x^3(a>0), at the points of intersection other than the origin.
1
Expert's answer
2020-05-21T17:40:06-0400

Area between two curves f and g is given by:

A = ab(f(y)g(y))dy\int^b_a (f(y) -g(y)) dy where a<y<b


Here,

g = y2=ax

x=y2ax= \frac{y^2}{a}


f = ay2=x3

x=a1/3y2/3x = a^{1/3} y^{2/3}



A = aa(y2a+a1/3y2/3)dy\int^a_{-a} (-\frac{y^2}{a} +a^{1/3} y^{2/3}) dy


=aay2ady+aaa1/3y2/3dy=\int^a_{-a} -\frac{y^2}{a} dy + \int^a_{-a} a^{1/3} y^{2/3} dy


=y33aaa+3y5/3a1/35aa=\left. \frac{-y^3}{3a} \right|^{a}_{-a} + \left. \frac{3y^{5/3}a^{1/3}}{5} \right|^{a}_{-a}


=a23+3a25a23+3a25-\frac{a^2}{3} + \frac{3a^2}{5} -\frac{a^2}{3} + \frac{3a^2}{5}


= 2a23+6a25-\frac{2a^2}{3} + \frac{6a^2}{5}


= 10a2+18a215\frac{-10a^2+18a^2}{15}


=8a215\frac{8a^2}{15}


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