Surface Integral Over a Half-Cylinder: 04-BS-1, December 2016, Q8

Question 8: Surface Integral Over a Half-Cylinder 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A half-cylinder of radius $2$, height $0\le z\le4$, restricted to $y\ge0$; integrand $x^2yz$.

The shaded half-tube ($y\ge0$, i.e. $\theta\in[0,\pi]$) of the cylinder $x^2+y^2=4$ between $z=0$ and $z=4$.

Find. $\displaystyle\iint_S x^2yz\,dS$.

Approach. Parametrize the cylinder by $(\theta,z)$, compute the surface-element factor $|\mathbf r_\theta\times\mathbf r_z|$, and integrate the substituted integrand over $\theta\in[0,\pi]$ (the $y\ge0$ half), $z\in[0,4]$.

Parametrize. $\mathbf r(\theta,z)=(2\cos\theta,\,2\sin\theta,\,z)$, $\theta\in[0,\pi]$ (since $y=2\sin\theta\ge0$ there), $z\in[0,4]$.
Surface element. $\mathbf r_\theta=(-2\sin\theta,2\cos\theta,0)$, $\mathbf r_z=(0,0,1)$, so $\mathbf r_\theta\times\mathbf r_z=(2\cos\theta,2\sin\theta,0)$, with magnitude $|\mathbf r_\theta\times\mathbf r_z|=2$. Thus $dS=2\,d\theta\,dz$.
Substitute the integrand. $x^2yz=(2\cos\theta)^2(2\sin\theta)(z)=8z\cos^2\theta\sin\theta$, so $$\iint_S x^2yz\,dS=\int_0^4\!\!\int_0^\pi 8z\cos^2\theta\sin\theta\cdot2\,d\theta\,dz=16\int_0^4z\,dz\int_0^\pi\cos^2\theta\sin\theta\,d\theta.$$
Evaluate each factor. $\displaystyle\int_0^4z\,dz=8$. For the angular integral, substitute $u=\cos\theta$, $du=-\sin\theta\,d\theta$ (limits $u:1\to-1$): $$\int_0^\pi\cos^2\theta\sin\theta\,d\theta=\int_{-1}^1u^2\,du=\frac23.$$
Combine. $$\iint_S x^2yz\,dS=16\cdot8\cdot\frac23=\boxed{\dfrac{256}{3}}\approx85.33$$

Quantity	Result
Surface element	$dS=2\,d\theta\,dz$
$\int_0^4z\,dz$	$8$
$\int_0^\pi\cos^2\theta\sin\theta\,d\theta$	$2/3$
$\iint_S x^2yz\,dS$	$256/3\approx85.33$

Concept & Theory

Topic. Parametrized surface integrals over a cylindrical surface.

Key relations. For a parametrization $\mathbf r(u,v)$, $\displaystyle\iint_S f\,dS=\iint_D f(\mathbf r(u,v))\,|\mathbf r_u\times\mathbf r_v|\,du\,dv$. For a right circular cylinder of radius $a$, the surface-element factor $|\mathbf r_\theta\times\mathbf r_z|=a$ is constant, so $dS=a\,d\theta\,dz$ — no extra weighting depending on $\theta$ or $z$.

Why this works. Substituting the cylinder's parametrization turns $x,y$ into $2\cos\theta,2\sin\theta$, converting the surface integral into an ordinary double integral over a rectangle $[0,\pi]\times[0,4]$ in $(\theta,z)$; the integrand separates into a pure-$z$ factor and a pure-$\theta$ factor, so the double integral factors into a product of two single integrals.

Common pitfall. Using $\theta\in[0,2\pi]$ (the full cylinder) instead of restricting to $[0,\pi]$ for the stated $y\ge0$ half; forgetting the constant factor $a=2$ from $|\mathbf r_\theta\times\mathbf r_z|$, i.e. computing $\iint x^2yz\,d\theta\,dz$ instead of $dS$; sign or substitution errors in the $u=\cos\theta$ trig integral.

Source. Stewart, Calculus: Early Transcendentals, Ch. 16 (§16.6 Parametric Surfaces and Their Areas; §16.7 Surface Integrals).

Question 8 of 8: Surface Integral Over a Half-Cylinder

Question 8: Surface Integral Over a Half-Cylinder 20 marks