## Thomas' Calculus 13th Edition

a) $2\sqrt 3$ b)$8$
a ) STEP 1) $A(x)$ = $\frac{1}{2}(side)(side)(sin\frac{\pi}{3})$ =$\frac{1}{2}(2\sqrt {sinx}(2\sqrt {sinx})(sin\frac{\pi}{3})$ =$\sqrt 3 \sin x$ STEP 2) $a$=$0$ , $b$=$\pi$ STEP 3 ) $V$=$\int^b_aA(x) dx$ =$\sqrt 3\int^\pi_0\sin x dx$ =$[-\sqrt 3\ cosx]^\pi_0$ =$\sqrt 3(1+1)$ =$2\sqrt 3$ b) STEP 1 ) $A(x)$ = $(side)^2$ = $(2\sqrt {\ sinx})((2\sqrt {\ sinx})$ =$4 \sin x$ STEP 2) $a$=$0$, $b$=$\pi$ STEP 3) $V$=$\int^b_aA(x)dx$ = $\int^\pi_04 \sin x \ dx$ =$[-4 \ cos x ]^\pi_0$ = $8$