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Consider the curve x=eʸ +e−ʸ /2

Now imagine we rotate it around the y-axis. Find the surface area that is generated by this rotation from y=0 to y=ln(2)

User MaiKaY
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Final answer:

The question pertains to calculating the surface area of a solid of revolution formed by rotating the curve x = e^y + e^-y / 2 around the y-axis, from y=0 to y=ln(2). This involves calculus, specifically integration and finding the derivative of x with respect to y. The formula 2πx ∙ ds is used, where ds is the arc length element.

Step-by-step explanation:

The student in question is asking about finding the surface area of a solid of revolution. Specifically, they want to calculate the surface area for the curve x = ey + e-y / 2 when it is revolved around the y-axis, between the limits of y=0 and y=ln(2). This is a typical calculus problem involving the use of integrals to find surface areas of solids of revolutions.

To solve this, we would typically use the formula for the surface area S of a solid of revolution about the y-axis, which is:

S = ∫ 2πx ∙ ds,

where x is the radius at a given y, and ds is an infinitesimally small arc length element of the curve over the interval of integration. To find ds, we need to use the formula ds = √(1 + (dx/dy)2) dy, where dx/dy is the derivative of x with respect to y.

After calculating the derivative dx/dy, you would integrate the function from y=0 to y=ln(2). The integral would give the total surface area of the solid of revolution.

User Chomp
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