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A cross-section perpendicular to the rotation axis is a disk of radius $\sin x$. Disc integration therefore givesThe result is in cubic coordinate units; the full rotation is already accounted for by the disk area.

Use completing the square after taking out the leading coefficient:Thus $a=3$, $b=-2$ and $c=-1$. Since a real square is nonnegative, the global minimum is $-1$, attained at $x=2$.

The derivative is $dy/dx=3x^2$, giving slope twelve at $x=2$. The point on the curve is $(2,9)$. The point-slope equation of the tangent line is consequently

Write $z=x+iy$. The complex modulus condition is equivalent to $(x-1{)}^2+y^2=4$. Thus the locus is a circle of radius two centred at $(1,0)$ in the complex plane. Its extreme points are $(-1,0)$, $(3,0)$, and $(1,\pm 2)$.

This is a separable differential equation. Multiply by $e^{-y}$ and integrate:The initial condition gives $C=-1$. HenceFor a real solution the outer logarithm requires a positive argument. The maximal interval containing the initial point is therefore $\boxed{|x|<\sqrt{e^2-1}}$. At either endpoint $y$ diverges, so the real solution cannot continue through it with a finite value.

For $f(x)=\sqrt{x}$, the first two derivatives at one are $f^{\prime }(1)=1/2$ and $f^{\prime \prime }(1)=-1/4$. The Taylor expansion beginsAll three displayed terms are nonzero terms of the expansion about one.

Rationalize with the complex conjugate $4-3i$:Thus the real part and imaginary part are