Question

Let R denote the region in the first quadrant bounded by the curve y=2−e x and the x and y axes. Which of the expressions below computes the surface area of the solid obtained by rotating the region R around the x-axis

A ∫ 0/ln2 2π(2−e x ) √1−e 2x dx
B ∫ 0/1 2π(2−e x ) √1−e 2x dx
C ∫ 0/ln2 2π(2−e x ) √1+2e 2x dx
D ∫ 0/ln2 2π(2−e x ) √1+e 2x dx
E ∫ 0/1 2π(2−e x ) √1+2e 2x dx
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Answer 1

The surface area of the solid created by rotating region R around the x-axis is calculated using the integral from 0 to ln(2) of 2π(2 - e^x) sqrt(1 + e^(2x)) dx, which corresponds to option D.

Step-by-step explanation:

To calculate the surface area of the solid obtained by rotating the region R around the x-axis, we'll use the formula for the surface area of a solid of revolution, which involves integrating 2π times the radius of rotation times the arc length. The radius of rotation is y, which is 2 - ex, and the arc length differential (sqrt(1 + (dy/dx)2)) for the curve y=2 - ex is sqrt(1 + e2x) because (dy/dx) is -ex. Now, we must determine the correct limits of integration. Notice that the curve intersects the x-axis when y = 0, which implies 2 - ex = 0, or x = ln(2). Hence, we integrate from 0 to ln(2).

Combining these, the integral for the surface area is ∫ 0 to ln(2) 2π(2 - ex) sqrt(1 + e2x) dx, which closely matches option D from the initial question. Clearly, the correct choice is D.