Worked examples · Sheet 2

Areas of Cartesian surfaces

Theory used: Area of a Cartesian surface Area of a surface

Exercise 1

Compute the area of the portion of the plane $z = 2 x + y$ that projects onto the triangle $T$ with vertices $(0, 0)$ , $(1, 0)$ , $(0, 1)$ .

f (x, y) = 2 x + y, f_{x} = 2, f_{y} = 1

1 + f_{x}^{2} + f_{y}^{2} = 1 + 4 + 1 = 6

Area = \iint_{T} 6 d x d y = 6 \cdot \frac{1}{2} = \frac{6}{2}

The surface is the graph of $f (x, y) = 2 x + y$ above the triangle $T$ , so the formula for the area of a Cartesian surface applies. The partial derivatives are constant, therefore the integrand $1 + f_{x}^{2} + f_{y}^{2} = 6$ is constant as well and the double integral is the product of $6$ and the area of $T$ . The triangle has legs of length $1$ , so its area is $\frac{1}{2}$ and the area of the portion of the plane is $\frac{6}{2}$ .

Exercise 2

Compute the area of the portion of the paraboloid

z = \frac{x ^{2} + y ^{2}}{2}

that projects onto the disk $x^{2} + y^{2} \leq 1$ .

f_{x} = x, f_{y} = y, 1 + f_{x}^{2} + f_{y}^{2} = 1 + x^{2} + y^{2}

Area = \iint_{x^{2} + y^{2} \leq 1} 1 + x^{2} + y^{2} d x d y = \int_{0}^{2 π} \int_{0}^{1} 1 + ρ^{2} ρ d ρ d θ

Area = 2 π [\frac{1}{3} (1 + ρ^{2})^{\frac{3}{2}}]_{0}^{1} = \frac{2 π}{3} (22 - 1)

The partial derivatives of $f (x, y) = \frac{x ^{2} + y ^{2}}{2}$ are $f_{x} = x$ and $f_{y} = y$ , so the area integrand is $1 + x^{2} + y^{2}$ . Since the domain is a disk and the integrand depends only on $x^{2} + y^{2}$ , it is convenient to pass to polar coordinates, in which $x^{2} + y^{2} = ρ^{2}$ and the area element becomes $ρ d ρ d θ$ . The integral in $θ$ gives the factor $2 π$ . The integral in $ρ$ has integrand $ρ 1 + ρ^{2}$ , which is the derivative of $\frac{1}{3} (1 + ρ^{2})^{\frac{3}{2}}$ , so an antiderivative is $\frac{1}{3} (1 + ρ^{2})^{\frac{3}{2}}$ . Evaluating between $0$ and $1$ we obtain $\frac{2 2 - 1}{3}$ , so the area is $\frac{2 π}{3} (22 - 1)$ .

Exercise 3

Compute the area of the graph of

f (x, y) = \frac{2}{3} (x^{\frac{3}{2}} + y^{\frac{3}{2}})

over the square $Q = [0, 1] \times [0, 1]$ .

f_{x} = x, f_{y} = y, 1 + f_{x}^{2} + f_{y}^{2} = 1 + x + y

\int_{0}^{1} 1 + x + y d x = [\frac{2}{3} (1 + x + y)^{\frac{3}{2}}]_{x = 0}^{x = 1} = \frac{2}{3} ((2 + y)^{\frac{3}{2}} - (1 + y)^{\frac{3}{2}})

Area = \frac{2}{3} \int_{0}^{1} ((2 + y)^{\frac{3}{2}} - (1 + y)^{\frac{3}{2}}) d y = \frac{2}{3} \cdot \frac{2}{5} [(2 + y)^{\frac{5}{2}} - (1 + y)^{\frac{5}{2}}]_{0}^{1} = \frac{4}{15} (3^{\frac{5}{2}} - 2^{\frac{5}{2}} - 2^{\frac{5}{2}} + 1) = \frac{4}{15} (93 - 82 + 1)

The partial derivatives are $f_{x} = x$ and $f_{y} = y$ , since the derivative of $\frac{2}{3} x^{\frac{3}{2}}$ is $x^{\frac{1}{2}}$ . The area integrand becomes $1 + x + y$ , and the function $f$ of the exercise is constructed precisely so that the squares of the derivatives under the root give a linear expression, which can be integrated in elementary terms. We integrate first in $x$ with $y$ constant: an antiderivative of $(1 + x + y)^{\frac{1}{2}}$ with respect to $x$ is $\frac{2}{3} (1 + x + y)^{\frac{3}{2}}$ . We then integrate the result in $y$ , with antiderivatives of the form $\frac{2}{5} (c + y)^{\frac{5}{2}}$ . In the final evaluation $3^{\frac{5}{2}} = 93$ and $2^{\frac{5}{2}} = 42$ appear, the latter twice, therefore the area equals $\frac{4}{15} (93 - 82 + 1)$ .