Question
Let R be the shaded region bounded by the graph of y = xex 2 , the line y = -2x , and the vertical line x = 1, as shown in the figure above.
(a) Find the area of R.
(b) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = -2.
(c) Write, but do not evaluate, an expression involving one or more integrals that gives the perimeter of R.
Answer/Explanation
Ans:
(a) Area = \(\int_{0}^{1}\left ( xe^{x^{2}}-(-2x) \right )dx\)
\(=\left [ \frac{1}{2}e^{x^{2}}+x^{2} \right ]_{x=0}^{x=1}\)
\(=\left ( \frac{1}{2}e+1 \right )-\frac{1}{2}=\frac{e+1}{2}\)
(b) Volume = \(\pi \int_{0}^{1}\left [ \left ( xe^{x^{2}}+2 \right )^{2} -(-2x+2)^{2}\right ]dx\)
(c) \(y’=\frac{d}{dx}\left ( xe^{x^{2}} \right )=e^{x^{2}}+2x^{2}e^{x^{2}}=e^{x^{2}}(1+2x^{2})\)
Perimeter = \(\sqrt{5}+2+e+\int_{0}^{1}\sqrt{1+\left [ e^{x^{2}}(1+2x^{2}) \right ]^{2}}dx\)