Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 91

$2$

Here, we have $A=\int_p^q [f(y)-g(y)] dy$ This implies that $A=(3)\int_{0}^{\pi/2} \sin y\sqrt {\cos y }dy-0 =(3)\int_{0}^{\pi/2} \sin y\sqrt {\cos y }dy$ Plug $p=\cos y \implies dp=-\sin y$ Then, we get $A=(-3)[\dfrac{2}{3}[\cos y)^{3/2}]_{0}^{\pi/2}=-2(0-1)=2$