Question
Consider the two regions, \(R_{1}\) and \(R_{2}\), shown in the figure below
(A) Is there a value of a > 0 that makes \(R_{1}\) and \(R_{2}\) have equal area? Justify your response.
(B) If the line x = b divides the region \(R_{1}\) into two regions of equal area, express b in terms of a.
(C) Express in terms of a the volume of the solid obtained by revolving the region \(R_{1}\) about the y-axis.
(D) If \(R_{2}\)is the base of a solid whose cross sections perpendicular to the y-axis are squares, find the volume of the solid in terms of a.
Answer/Explanation
(A) Let \(A_{i}\) denote the area of \(R_{i}\).Then,
\(A_{1}=\int_{0}^{a^{2}}\sqrt{x}dx=\left [ \frac{2}{3}x^{3/2} \right ]^{a^{2}}_{0}=\frac{2a^{3}}{3}\)
and
\(A_{2}=\int_{0}^{a}y^{2}dy=\left [ \frac{y^{3}}{3} \right ]^{a}_{0}=\frac{a^{3}}{3}\)
Therefore, you solve \(A_{1}=A_{2}\) for a:
\(\frac{2a^{3}}{3}=\frac{a^{3}}{3}\Leftrightarrow a^{3}=0\Leftrightarrow a=0\)
Thus, there are no solutions a > 0 such that \(A_{1}=A_{2}\)
(B)\(R_{1}\) is divided into two regions of equal area, each expressed by the following integral equations:
\(\frac{A_{1}}{2}=\int_{0}^{b}\sqrt{x}dx=\frac{2}{3}b^{3/2}\)
ans
\(\frac{A_{1}}{2}=\int_{b}^{a^{2}}\sqrt{x}dx=\frac{2}{3}a^{3}-\frac{2}{3}b^{3/2}\)
Upon setting these two expressions equal, you solve for b:
\(\frac{2}{3}b^{3/2}=\frac{2}{3}a^{3}-\frac{2}{3}b^{3/2}\Leftrightarrow 2b^{3/2}=a^{3}\Leftrightarrow b=\frac{a^{2}}{\sqrt[3]{4}}\)
(C)The solid obtained by rotating R1 about the y-axis has cross sections of washers with inner radius \(x=y^{2}\) and outer radius \(x=a^{2}\). Therefore, the volume is expressed as
\(\pi \int_{0}^{a}(a^{4}-y^{4})dy=\pi \left [ a^{4}y-\frac{1}{5}y^{5} \right ]^{a}_{0}=\pi \left ( \frac{5a^{2}}{5}-\frac{a^{5}}{5} \right )=\frac{4\pi a^{5}}{5}\)
(D)At each y belonging to [0, a], the area of a square cross section is given by \(A(y)=(y^{2})^{2}=y^{4}\).Therefore, the volume of the solid is just the value of the integral:
\(\int_{0}^{a}y^{4}dy=\frac{1}{5}a^{5}\)