this question is designed to be answered without a calculator. region r is bounded by the curves y = 4x² and…

Answer:

C. $\pi$

Explanation:

Step1: Find intersection points

Set $4x^{2}=4$, then $x^{2} = 1$, so $x=- 1,1$. Solve $y = 4x^{2}$ for $x$, we get $x=\pm\frac{\sqrt{y}}{2}$. The diameter $d$ of the semi - circle cross - section perpendicular to the $y$ - axis is $d = 2\times\frac{\sqrt{y}}{2}=\sqrt{y}$ (since we consider the distance between the two $x$ values for a given $y$). The radius $r$ of the semi - circle is $r=\frac{d}{2}=\frac{\sqrt{y}}{2}$.

Step2: Find area formula of cross - section

The area formula for a semi - circle is $A(y)=\frac{1}{2}\pi r^{2}$. Substitute $r = \frac{\sqrt{y}}{2}$ into it, we have $A(y)=\frac{1}{2}\pi(\frac{\sqrt{y}}{2})^{2}=\frac{\pi y}{8}$.

Step3: Set up integral for volume

The limits of integration for $y$ are from $0$ to $4$. Using the formula for the volume of a solid with known cross - sectional area $V=\int_{a}^{b}A(y)dy$, we get $V=\int_{0}^{4}\frac{\pi y}{8}dy$.

Step4: Evaluate the integral

$V=\frac{\pi}{8}\int_{0}^{4}y;dy$. By the power rule $\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\frac{\pi}{8}\times\frac{y^{2}}{2}\big|_{0}^{4}=\frac{\pi}{16}(4^{2}-0^{2})=\pi$.