Calculus III Advanced (Course) (14.02) (Homework)

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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

$$\int_{R} \int x^{2}y^{2} \:dA $$ $$= \int_{0}^{4} \int_{1/2(y)}^{\sqrt{y}} x^{2}y^{2} \:dx \:dy\:\:\:\: \color{red}{ \text{ Apply Fubini's Theorem}}$$
$$= \left. \frac{1}{3} \int_{0}^{4} x^{3}y^{2} \right]_{1/2(y)}^{\sqrt{y}} \: dy $$
$$= \frac{1}{3} \int_{0}^{4} \left( y^{7/2}- \frac{1}{8}y^{5} \right) \: dy $$
$$= \frac{1}{3} \left[ \frac{2}{9} y^{9/2}- \frac{1}{48}y^{6} \right]_{0}^{4} = \frac{256}{27} $$
Figure 1

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

\( \frac{1}{2}\pi (2)^{2} = 2 \pi\)
Figure 2

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.

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.

Figure 4

\(2x\) \(= x^{2}+y^{2}\)
\(y\) \(= \sqrt{2x-x^{2}}\)
$$Volume $$ $$=\int_{0}^{4} \int_{-\sqrt{2x-x^{2}}}^{\sqrt{2x-x^{2}}} x^{2}+y^{2} \: dy \: dx - \int_{0}^{4} \int_{-\sqrt{2x-x^{2}}}^{\sqrt{2x-x^{2}}} 2x \: dy \: dx$$

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

\(Region: \: 0 \leqslant y \leqslant 3, \: y/3 \leqslant x \leqslant 1 \)
Figure 5

$$\int_{0}^{1} \int_{0}^{3x} \frac{1}{1+x^{4}} \: dy \: dx $$ $$= \frac{3}{8}\pi $$

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

\( 0 \leqslant y \leqslant x \).
$$Average \: Value $$ $$= \frac{1}{A} \int_{R} \int f(x,y) \:dA$$
$$=\frac{1}{\frac{1}{2}} \int_{0}^{1} \int_{0}^{x} \frac{1}{x+y} \: dy \: dx $$
$$= 2 \int_{0}^{1} \left[ \ln | x+y| \right]_{0}^{x} \: dx $$
$$= 2 \int_{0}^{1} ( \ln(2x) - \ln(x)) \: dx $$
$$= 2 \int_{0}^{1} \ln 2 \: dx = 2 \ln 2$$

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Parent Article: Calculus III Advanced (Course)