We follow Definition 13.5.4. We start by noting that \(f_x(x,y) = -1\) and \(f_y(x,y) = -2\text{.}\) To define \(R\text{,}\) we use bounds \(0\leq y\leq 2-x/2\) and \(0\leq x\leq 4\text{.}\) Therefore \begin{align*} S \amp = \iint_R\ dS\\ \amp = \int_0^4\int_0^{2-x/2} \sqrt{1+(-1)^2+(-2)^2}\ dy\ dx\\ \amp = \int_0^4 \sqrt{6}\left(2-\frac x2\right)\ dx\\ \amp = 4\sqrt{6}. \end{align*}

Because the surface is a triangle, we can figure out the area using geometry. Considering the base of the triangle to be the side in the \(x\)-\(y\) plane, we find the length of the base to be \(\sqrt{20}\text{.}\) We can find the height using our knowledge of vectors: let \(\vec u\) be the side in the \(x\)-\(z\) plane and let \(\vec v\) be the side in the \(x\)-\(y\) plane. The height is then \(\norm {\vec u - \proj{u}{v}} = 4\sqrt{6/5}\text{.}\) Geometry states that the area is thus \begin{equation*} \frac 12\cdot4\sqrt{6/5}\cdot\sqrt{20} = 4\sqrt{6}. \end{equation*}

We affirm the validity of our formula.

We start by computing partial derivatives and find \begin{equation*} f_x(x,y) = \frac{-x}{\sqrt{a^2-x^2-y^2}} \text{ and } f_y(x,y) = \frac{-y}{\sqrt{a^2-x^2-y^2}}. \end{equation*}

As our function \(f\) only defines the top upper hemisphere of the sphere, we double our surface area result to get the total area: \begin{align*} S \amp = 2\iint_R \sqrt{1+ f_x(x,y)^2+f_y(x,y)^2}\ dA\\ \amp = 2\iint_R \sqrt{1+ \frac{x^2+y^2}{a^2-x^2-y^2}}\ dA. \end{align*}

The region \(R\) that we are integrating over is the circle, centered at the origin, with radius \(a\text{:}\) \(x^2+y^2=a^2\text{.}\) Because of this region, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions \(x=r\cos(\theta)\text{,}\) \(y=r\sin(\theta)\text{,}\) \(dA = r\ dr\ d\theta\) and bounds \(0\leq\theta\leq2\pi\) and \(0\leq r\leq a\text{,}\) we have: \begin{align*} S \amp = 2\int_0^{2\pi}\int_0^a \sqrt{1+\frac{r^2\cos^2(\theta) +r^2\sin^2(\theta) }{a^2-r^2\cos^2(\theta) -r^2\sin^2(\theta) }}\ r\ dr\ d\theta\\ \amp =2\int_0^{2\pi}\int_0^ar\sqrt{1+\frac{r^2}{a^2-r^2}}\ dr\ d\theta\\ \amp =2\int_0^{2\pi}\int_0^ar\sqrt{\frac{a^2}{a^2-r^2}}\ dr\ d\theta.\\ \end{align*} Apply substitution \(u=a^2-r^2\) and integrate the inner integral, giving \begin{align*} \amp = 2\int_0^{2\pi} a^2\ d\theta\\ \amp = 4\pi a^2. \end{align*}

Our work confirms our previous formula.

We begin by computing partial derivatives. \begin{equation*} f_x(x,y) = -\frac{xh}{a\sqrt{x^2+y^2}} \text{ and } -\frac{yh}{a\sqrt{x^2+y^2}}. \end{equation*}

Since we are integrating over the circle \(x^2+y^2=a^2\text{,}\) we again use polar coordinates. Using the standard substitutions, our integrand becomes \begin{equation*} \sqrt{1+ \left(\frac{hr\cos(\theta) }{a\sqrt{r^2}}\right)^2 + \left(\frac{hr\sin(\theta) }{a\sqrt{r^2}}\right)^2}. \end{equation*}

This may look intimidating at first, but there are lots of simple simplifications to be done. It amazingly reduces to just \begin{equation*} \sqrt{1+\frac{h^2}{a^2}} = \frac1a\sqrt{a^2+h^2}. \end{equation*}

Our polar bounds are \(0\leq\theta\leq2\pi\) and \(0\leq r\leq a\text{.}\) Thus \begin{align*} S \amp = \int_0^{2\pi}\int_0^ar\frac1a\sqrt{a^2+h^2}\ dr\ d\theta\\ \amp = \int_0^{2\pi} \left.\left(\frac12r^2\frac1a\sqrt{a^2+h^2}\right)\right|_0^ad\theta\\ \amp = \int_0^{2\pi} \frac12a\sqrt{a^2+h^2} \ d\theta\\ \amp = \pi a\sqrt{a^2+h^2}. \end{align*}

This matches the formula found in the back of this text.

It is straightforward to compute \(f_x(x,y) = 2x\) and \(f_y(x,y) = -3\text{.}\) Thus the surface area is described by the double integral \begin{equation*} \iint_R \sqrt{1+(2x)^2+(-3)^2}\ dA = \iint_R \sqrt{10+4x^2}\ dA. \end{equation*}

As with integrals describing arc length, double integrals describing surface area are in general hard to evaluate directly because of the square–root. This particular integral can be easily evaluated, though, with judicious choice of our order of integration.

Integrating with order \(dx\ dy\) requires us to evaluate \(\int \sqrt{10+4x^2}\ dx\text{.}\) This can be done, though it involves Integration By Parts and \(\sinh^{-1}(x)\text{.}\) Integrating with order \(dy\ dx\) has as its first integral \(\int \sqrt{10+4x^2}\ dy\text{,}\) which is easy to evaluate: it is simply \(y\sqrt{10+4x^2}+C\text{.}\) So we proceed with the order \(dy\ dx\text{;}\) the bounds are already given in the statement of the problem. \begin{align*} \iint_R\sqrt{10+4x^2}\ dA \amp = \int_0^4\int_{-x}^x\sqrt{10+4x^2}\ dy \ dx\\ \amp = \int_0^4\left.\big(y\sqrt{10+4x^2}\big)\right|_{-x}^x dx\\ \amp =\int_0^4\big(2x\sqrt{10+4x^2}\big)\ dx.\\ \end{align*} Apply substitution with \(u = 10+4x^2\text{:}\) \begin{align*} \amp = \left.\left(\frac16\big(10+4x^2\big)^{3/2}\right)\right|_0^4\\ \amp = \frac13\big(37\sqrt{74}-5\sqrt{10}\big) \approx 100.825\text{u} ^2. \end{align*}

So while the region \(R\) over which we integrate has an area of 16u\(^2\text{,}\) the surface has a much greater area as its \(z\)-values change dramatically over \(R\text{.}\)