Find the area of the part of hyperbolic paraboloid that lies between the cylinders and .

Question

Find the area of the part of hyperbolic paraboloid  that lies between the cylinders  and  .

Quizplus · Accepted Answer

The cylinders given by the equations
$$y^2+z^2=1$$
and
$$y^2+z^2=4$$
are centered at the origin and have radii 1 and 2, respectively. Thus, the hyperbolic paraboloid $x=y^2-z^2$ intersects these cylinders in the curves
$$y^2+z^2=1,\quad x=1-y^2,$$
and
$$y^2+z^2=4,\quad x=4-y^2.$$
The region between these two curves consists precisely of the portion of the hyperbolic paraboloid with $1-y^2\leq x\leq 4-y^2$. This is a geometrically evident explanation, but we can also see it by solving for $y$ in terms of $x$,
$$y=\pm\sqrt{1-x},\quad y=\pm\sqrt{4-x},$$
which gives $-1\leq y\leq 1$ and $-\sqrt{3}\leq y\leq\sqrt{3}$ for $1\leq x\leq 4$, respectively. Therefore, the region in question is described by the inequalities
$$-1\leq y\leq 1,\quad 1-y^2\leq x\leq 4-y^2.$$
The area of the region between two curves $y=f(x)$ and $y=g(x)$ over an interval $[a,b]$ is given by
$$\int_a^b|f(x)-g(x)|\,dx,$$
so the area in question is
$$\int_{-1}^1|(1-y^2)-(4-y^2)|\,dy=\int_{-1}^1 3\,dy=\boxed{6}.$$

Find the Area of the Part of Hyperbolic Paraboloid