Question
Let f(x) = e2x . Let R be the region in the first quadrant bounded by the graph of f, the coordinate axes, and the vertical line x = k, where k > 0. The region R is shown in the figure above.
(a) Write, but do not evaluate, an expression involving an integral that gives the perimeter of R in terms of k.
(b) The region R is rotated about the x-axis to form a solid. Find the volume, V, of the solid in terms of k.
(c) The volume V, found in part (b), changes as k changes. If \(\frac{dk}{dt}=\frac{1}{3}, determine \frac{dV}{t}when k=\frac{1}{2}.\)
Answer/Explanation
Ans:
(a)
\(\frac{dy}{dx}=2e^{2x}\)
\(p = 1 + k + e^{2x}+ \int_{0}^{k}\sqrt{1+(2e^{2x})^{2}}dx\)
(b)
u = 4x
\(\frac{du}{dx}=4\)
\(v = \int_{0}^{k}\pi r^{2}dx=\int_{0}^{k}\pi (e^{2x})^{2}dx=\pi \int_{0}^{k}e^{4x}dx\)
\(= \pi \int_{0}^{k}\frac{1}{4}e^{4}\cdot du=\frac{\pi }{4}\left [ e^{4x} \right ]_{0}^{k}=\frac{\pi }{4}\left ( e^{4k} -e^{4(0)}\right )\)
\(=\frac{\pi }{4}e^{4k}-1\left ( \frac{\pi }{4} \right )\)
\(v=\frac{\pi }{4}e^{4k}- \frac{\pi }{4}\)
(c)
\(\frac{dV}{dt}=\frac{dV}{dk}\cdot \frac{dk}{dt}=\frac{d}{dk}\left ( \frac{\pi }{4}e^{4k}-\frac{\pi }{4} \right )\cdot \left ( \frac{1}{3} \right )=\frac{\pi }{4}(4)e^{4k}\cdot \left ( \frac{1}{3} \right )\)
\(\frac{dV}{dt}=\frac{\pi }{3}e^{4k}|_{k=\frac{1}{2}}=\frac{\pi }{3}\cdot e^{2}\)