calculate the lateral surface area of the solids resulting from rotating the following curves about the x-axis f(x)=sinx on the open interval...

Use the formula:

S=Integral
2pi y ds, y=f(x)=sinx

S=Integral 2pi sin x
ds

Calculate ds
with:

ds=sqrt(1+(dy/dx)^2)

calculate
dy/dx=cosx => ds=sqrt(1+(cos x)^2)

Calculate
S:

S = Integral 2pi sin xsqrt(1+(cos x)^2)
dx

Let (cos x)=k=> (cos x)'=k' => -sin
x=dk/dx

S = -Integral 2pi sqrt(1+(k)^2) dk=-2pi Integral
sqrt(1+(k)^2)
dk

S=(-2pi/2)(ksqrt(1+(k)^2)+ln(k+sqrt(1+(k)^2)))

Do
the substitution k=cos x

S=pi(cos 0sqrt(1+(cos0)^2)+ln(cos
0+sqrt(1+(cos 0)^2)-cos (pi/2)sqrt(1+(cos(pi/2))^2)-ln(cos (pi/2)+sqrt(1+(cos
(pi/2))^2)))

S=pi(sqrt2+ln(1+sqrt2)-ln(0+sqrt1))

S=pi(sqrt2+ln(1+sqrt2)-ln(1))

S=pi(sqrt2+ln(1+sqrt2)-0)

S=pi(sqrt2+ln(1+sqrt2))

Answer:
the lateral surface of the solid of
revolution:S=pi(sqrt2+ln(1+sqrt2))