Parameter-dependent integrals and applications

Parameter differentiation as a tool. For fixed limits $a,b$ independent of $\alpha$, Leibnitz's rule states

where: $f(x,\alpha)$ is the integrand depending on both the integration variable $x$ and the parameter $\alpha$, and the right side differentiates inside the integral, treating $x$ as fixed. The strategy: pick a parameter that makes $\partial f/\partial\alpha$ far simpler than $f$ itself, integrate the simplified expression back with respect to $\alpha$, and fix the resulting constant of integration using one special value of $\alpha$ (often $\alpha\to\infty$ or $\alpha=0$, where the original integral is either obviously zero or independently known).

Plane area via a double integral. For a plane region $R$,

with $dA = dx\,dy$ in Cartesian coordinates, or $dA = r\,dr\,d\theta$ in polar coordinates (Jacobian $J=r$) when $R$ is circular or sectorial — a disc, an annulus, a region bounded by $r=g(\theta)$.

Volume via a double or triple integral. For a solid of height $z=f(x,y)$ standing over a base region $R$,

while a genuinely three-dimensional solid $V$ uses $\text{Volume} = \iiint_V dV$, with $dV = r\,dr\,d\theta\,dz$ in cylindrical coordinates or $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ in spherical coordinates when the solid has that symmetry (a cylinder, a cone, a sphere or a spherical cap).

Surface area of $z=f(x,y)$ over $R$.

where: $f_x=\partial f/\partial x$ and $f_y=\partial f/\partial y$ are the surface's two partial slopes at each point, and the square root is the local "tilt factor" that stretches a flat patch $dA$ into the corresponding tilted patch of surface. It is the direct two-variable analogue of the arc-length formula $s=\int\sqrt{1+(y')^2}\,dx$: one slope becomes two, and length becomes area.

Problem. Find the surface area of the part of the paraboloid $z=x^2+y^2$ that lies above the disc $x^2+y^2\le 1$ (all lengths in metres).

Step 1. Compute the partial slopes: $f(x,y)=x^2+y^2$ gives $f_x=2x$, $f_y=2y$, so $1+f_x^2+f_y^2 = 1+4x^2+4y^2$, and the surface-area integral is

Step 2. Switch to polar coordinates, since $R$ is a disc and the integrand depends only on $x^2+y^2=r^2$: with $dA=r\,dr\,d\theta$, $0\le r\le1$, $0\le\theta\le2\pi$,

Step 3. Evaluate the inner integral by substituting $u=1+4r^2$, $du=8r\,dr$ (so $r=0\Rightarrow u=1$, $r=1\Rightarrow u=5$):

Step 4. The outer integral over $\theta$ just multiplies by $2\pi$, since the inner result is independent of $\theta$:

Answer: $S = \dfrac{\pi(5\sqrt5-1)}{6}\approx 5.330\text{ m}^2$.

Competitive stretch

The same disc also bounds a solid: the volume trapped under the paraboloid and above the disc is $V=\iint_R (x^2+y^2)\,dA = \int_0^{2\pi}\int_0^1 r^2\cdot r\,dr\,d\theta = 2\pi\left[\dfrac{r^4}{4}\right]_0^1 = \dfrac{\pi}{2}\approx1.571\text{ m}^3$. Comparing the two results is instructive: the surface area needed the tilt-factor square root, while the volume underneath is a plain height integral — same region, same polar setup, structurally different integrand.

• Forgetting to fix the constant of integration after integrating $I'(\alpha)$ back — the parameter-differentiation method is incomplete until $C$ is pinned down using a known value or limit of $I(\alpha)$.
• Applying the surface-area formula with only one partial derivative squared — it needs both $f_x^2$ and $f_y^2$ under the square root, unlike single-variable arc length which has just one.
• Using Cartesian $dA=dx\,dy$ for a circular region when polar $dA=r\,dr\,d\theta$ would make the integral far simpler — always look at the region's shape before choosing coordinates.
• Confusing "volume under a surface" ($\iint_R f\,dA$, a double integral) with "volume of a genuine 3D solid" ($\iiint_V dV$, a triple integral) — pick whichever matches how the solid is actually described.

Ask the tutor: if you are unsure whether a given problem wants parameter differentiation, an area/volume setup, or a surface-area setup — or which coordinate system to switch to — ask the AI tutor to walk through a few more mixed examples with you.