Calculus III Advanced (Course) (14.02) (Homework)
Contents
- 1 Section 14.2 Homework
- 1.1 Exercise 14.2.8 Evaluating a Double Integral
- 1.2 Exercise 14.2.12 Evaluating a Double Integral
- 1.3 Exercise 14.2.18 Evaluating a Double Integral
- 1.4 Exercise 14.2.24 Finding Volume
- 1.5 Exercise 14.2.34 Volume for a Region Bounded by Two Surfaces
- 1.6 Exercise 14.2.48 Evaluating an Iterated Integral
- 1.7 Exercise 14.2.54 Average Value
- 2 Internal Links
Section 14.2 Homework
From Calculus 10e by Larson and Edwards, p. 983. Exercises 8, 12, 18, 24, 34, 48, 54.
Exercise 14.2.8 Evaluating a Double Integral
Sketch the region \(R\) and evaluate the iterated integral \(\int_{R} \int f(x,y) \:dA\).
- $$ \int_{0}^{4} \int_{1/2(y)}^{\sqrt{y}} x^{2}y^{2} \:dx \:dy$$
Solution The bounds are
- \(0 \leqslant y \leqslant 4\)
and
- \(1/2(y) \leqslant x \leqslant \sqrt{y}\).
The bounds for the sketch are \(x=y/2\) and \(x=\sqrt{y}\). produces
$$\int_{R} \int x^{2}y^{2} \:dA $$ | $$= \int_{0}^{4} \int_{1/2(y)}^{\sqrt{y}} x^{2}y^{2} \:dx \:dy\:\:\:\: \color{red}{ \text{ Apply Fubini's Theorem}}$$ |
$$= \left. \frac{1}{3} \int_{0}^{4} x^{3}y^{2} \right]_{1/2(y)}^{\sqrt{y}} \: dy $$ | |
$$= \frac{1}{3} \int_{0}^{4} \left( y^{7/2}- \frac{1}{8}y^{5} \right) \: dy $$ | |
$$= \frac{1}{3} \left[ \frac{2}{9} y^{9/2}- \frac{1}{48}y^{6} \right]_{0}^{4} = \frac{256}{27} $$ |
Exercise 14.2.12 Evaluating a Double Integral
Set up integrals for both integration orders. Use the more convenient order to evaluate the integral over the region \(R\).
- $$ \int_{R} \int \sin x \sin y \: dA $$
\(R\): rectangle with vertices \((-\pi,0), \: (\pi,0), \: (\pi, \pi/2),\:(-\pi,\pi/2)\)
Solution The region \(R\) is a rectangle bordered by the vertices, as shown in Figure 2, with an area equal to \(\pi\). The given bounds produce an area with zero size. Change the equation to
- $$ \int_{-\pi}^{\pi} \int_{0}^{\pi/2} \sin x \sin y \: dA $$
Now to integrate.
$$\int_{-\pi}^{ \pi} \int_{0}^{\pi/2} \sin x \sin y \: dy \:dx $$ | $$= \int_{-\pi}^{ \pi} \sin x [-\cos y]_{0}^{\pi/2} \:dx $$ | |
$$= \int_{-\pi}^{ \pi} (0)\sin x \:dx = 0$$ |
Exercise 14.2.18 Evaluating a Double Integral
Set up integrals for both orders of integration. Use the more convenient order to evaluate the integral over the region \(R\).
- $$ \int_{R} \int (x^{2}+y^{2}) \: dA$$
\(R\): semicircle bounded by \(y=\sqrt{4-x^{2}}, \: y=0\).
Solution The region \(R\) is a semi-circle as shown in Figure 2. The area is
- \( \frac{1}{2}\pi (2)^{2} = 2 \pi\)
Polar coordinates work best for this problem.
$$ \frac{1}{2} \int \int_{R} (x^{2}+y^{2}) \: dA$$ | $$=\frac{1}{2}\int_{0}^{\pi} \int_{0}^{2} r^{2} \:rd rd\theta $$ |
$$= \left. \int_{0}^{\pi} r^{4} \right]_{0}^{2} \:d \theta $$ | |
$$= \frac{1}{8} \int_{0}^{\pi} 16 \:d \theta$$ | |
$$= \left. \frac{1}{8} 16\theta \right]_{0}^{\pi} = 2 \pi $$ |
Exercise 14.2.24 Finding Volume
Use a double integral to find the volume for the solid in Figure 3.
$$ \int_{0}^{2} \int_{0}^{y} (4-y^{2}) \:dx \: dy $$ | $$= \left. \int_{0}^{2} (4-y^{2})x \right]_{0}^{y} \: dy $$ |
$$= \int_{0}^{2} (4-y^{2})y \: dy$$ | |
$$= \left. 4y-\frac{1}{3} y^{3} \right]_{0}^{2}$$ | |
$$= (4)(2)-\frac{1}{3} (2)^{3} - 4(0)-\frac{1}{3} (0)^{3} = \frac{16}{3}$$ |
Exercise 14.2.34 Volume for a Region Bounded by Two Surfaces
Set up a double integral to find the volume for the solid region bounded by the graphs shown in Figure 4. Do not evaluate the integral.
\(2x\) | \(= x^{2}+y^{2}\) |
\(y\) | \(= \sqrt{2x-x^{2}}\) |
$$Volume $$ | $$=\int_{0}^{4} \int_{-\sqrt{2x-x^{2}}}^{\sqrt{2x-x^{2}}} x^{2}+y^{2} \: dy \: dx - \int_{0}^{4} \int_{-\sqrt{2x-x^{2}}}^{\sqrt{2x-x^{2}}} 2x \: dy \: dx$$ |
Exercise 14.2.48 Evaluating an Iterated Integral
Sketch the integration region. Evaluate the iterated integral.
- $$\int_{0}^{3} \int_{y/3}^{1} \frac{1}{1+x^{4}} \: dx \: dy $$
Solution
- \(Region: \: 0 \leqslant y \leqslant 3, \: y/3 \leqslant x \leqslant 1 \)
$$\int_{0}^{1} \int_{0}^{3x} \frac{1}{1+x^{4}} \: dy \: dx $$ | $$= \frac{3}{8}\pi $$ |
Exercise 14.2.54 Average Value
Find the average value for \(f(x,y)\) over the plane region \(R\).
- $$f(x,y) = \frac{1}{x+y}$$
\(R\): triangle with vertices (0,0), (1,0), (1,1)
Solution The area for the triangle is 1/2. The bounds are
- \( 0 \leqslant x \leqslant 1 \)
and
- \( 0 \leqslant y \leqslant x \).
$$Average \: Value $$ | $$= \frac{1}{A} \int_{R} \int f(x,y) \:dA$$ |
$$=\frac{1}{\frac{1}{2}} \int_{0}^{1} \int_{0}^{x} \frac{1}{x+y} \: dy \: dx $$ | |
$$= 2 \int_{0}^{1} \left[ \ln | x+y| \right]_{0}^{x} \: dx $$ | |
$$= 2 \int_{0}^{1} ( \ln(2x) - \ln(x)) \: dx $$ | |
$$= 2 \int_{0}^{1} \ln 2 \: dx = 2 \ln 2$$ |
Internal Links
Parent Article: Calculus III Advanced (Course)