Worked examples · Sheet 2

Double integrals over normal domains

Theory used: Normal domain Reduction formulas on normal domains

Exercise 1

Compute the double integral

\iint_{D} x y d x d y, D = {(x, y) : 0 \leq x \leq 1, x^{2} \leq y \leq x} .

\iint_{D} x y d x d y = \int_{0}^{1} (\int_{x^{2}}^{x} x y d y) d x

\int_{x^{2}}^{x} x y d y = x [\frac{y ^{2}}{2}]_{y = x^{2}}^{y = x} = \frac{x ( x ^{2} - x ^{4} )}{2} = \frac{x ^{3} - x ^{5}}{2}

\int_{0}^{1} \frac{x ^{3} - x ^{5}}{2} d x = \frac{1}{2} (\frac{1}{4} - \frac{1}{6}) = \frac{1}{2} \cdot \frac{1}{12} = \frac{1}{24}

The domain is given directly in normal form with respect to the $x$ -axis: $x$ ranges over the interval $[0, 1]$ and, for each fixed $x$ , $y$ goes from the parabola $y = x^{2}$ to the line $y = x$ . The reduction formula therefore gives an inner integral in $y$ with limits $x^{2}$ and $x$ . In the inner integral the factor $x$ is constant and comes out of the integral sign; the antiderivative of $y$ is $\frac{y ^{2}}{2}$ and evaluating at the limits produces the polynomial $\frac{x ^{3} - x ^{5}}{2}$ . The integral of this polynomial over $[0, 1]$ equals $\frac{1}{24}$ .

Exercise 2

Compute the double integral

\iint_{T} x d x d y,

where $T$ is the triangle with vertices $(0, 0)$ , $(2, 0)$ and $(0, 2)$ .

T = {(x, y) : {0 \leq x \leq 2 0 \leq y \leq 2 - x}

\iint_{T} x d x d y = \int_{0}^{2} (\int_{0}^{2 - x} x d y) d x = \int_{0}^{2} x (2 - x) d x

\int_{0}^{2} (2 x - x^{2}) d x = [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3}

The first step is the description of the triangle in normal form. The slanted side joins $(2, 0)$ and $(0, 2)$ , so it lies on the line $x + y = 2$ , that is, $y = 2 - x$ . For each $x \in [0, 2]$ the vertical segment inside $T$ starts at the $x$ -axis and reaches the slanted line, therefore $0 \leq y \leq 2 - x$ . In the inner integral the function $x$ is constant with respect to $y$ , so the integral equals $x$ times the length of the segment, that is, $x (2 - x)$ . The integral of the polynomial $2 x - x^{2}$ over $[0, 2]$ equals $\frac{4}{3}$ .

Exercise 3

Compute the double integral

\iint_{D} y d x d y, D = {(x, y) : - 1 \leq y \leq 2, y^{2} \leq x \leq y + 2} .

\iint_{D} y d x d y = \int_{- 1}^{2} (\int_{y^{2}}^{y + 2} y d x) d y = \int_{- 1}^{2} y (y + 2 - y^{2}) d y

\int_{- 1}^{2} (y^{2} + 2 y - y^{3}) d y = [\frac{y ^{3}}{3} + y^{2} - \frac{y ^{4}}{4}]_{- 1}^{2} = (\frac{8}{3} + 4 - 4) - (- \frac{1}{3} + 1 - \frac{1}{4}) = \frac{8}{3} - \frac{5}{12} = \frac{27}{12} = \frac{9}{4}

The domain

D

, normal with respect to the

y

-axis: every horizontal segment enters through the parabola

x = y^{2}

and exits through the line

x = y + 2

The domain is normal with respect to the $y$ -axis: $y$ ranges over $[- 1, 2]$ and, for each fixed $y$ , $x$ goes from the parabola $x = y^{2}$ to the line $x = y + 2$ . The limits $- 1$ and $2$ are the ordinates of the intersection points of the two curves, solutions of $y^{2} = y + 2$ . Since the domain is described with respect to $y$ , the inner integral is in $x$ ; the function $y$ is constant with respect to $x$ , so the inner integral equals $y$ times the length of the horizontal segment, that is, $y (y + 2 - y^{2})$ . Expanding the product gives a polynomial in $y$ , whose integral over $[- 1, 2]$ equals $\frac{9}{4}$ . The description with respect to the $x$ -axis would have been more laborious, since the lower boundary would change expression at the point $x = 1$ and the domain would have to be split into two parts.