Let R be the region enclosed by $y = e^{2 x}, y = 0, and y = 2$ $y= e^(2x), y=0, and y=2$ . What is the volume of the solid produced by revolving R around the x-axis?

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Answer 1 · 2017-01-18T06:49:13

$\frac{3}{4} π = 2.3562$ $3/4pi=2.3562$ cubic units. nearly.

y = 2 meets $y = e^{2 x}$ $y = e^(2x)$ , at $(\frac{1}{2} ln 2, 2) .$ $(1/2ln 2, 2).$

The x-axis y = 0 is the asymptote to $y = e^{2 x} > 0$ $y = e^(2x)>0$ .

Now,

volume $= π \int y^{2} d x$ $= pi int y^2 dx$ , for x from $- \infty \to \frac{1}{2} ln 2$ $-oo to 1/2 ln 2$

$- π \int e^{4 x} d x$ $-pi int e^(4x) dx$ , for x from $- \infty \to \frac{1}{2} ln 2$ $-oo to 1/2 ln 2$

$= \frac{π}{4} [e^{4 x}]$ $=pi/4[e^(4x)]$ , between the limits

$= \frac{π}{4} [e^{2 ln 2} - e^{0}]$ $=pi/4[e^(2ln2)-e^0]$

$= \frac{π}{4} [4 - 1)$ $=pi/4[4-1)$ , using $e^{ln a} = a$ $e^ln a = a$ .

$= \frac{3}{4} π = 2.3562$ $=3/4pi=2.3562$ cubic units. nearly.

The graph shows the area that is revolved.

graph{(y-.1)(e^(2x)-y)(x-.5ln2+.0001y)=0 [-1.5, 1.5, 0, 2,1]}

Let R be the region enclosed by y=e2x,y=0,andy=2y= e^(2x), y=0, and y=2. What is the volume of the solid produced by revolving R around the x-axis?