Calculus III 14 Exam 1

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Chapter 14 Exam

From Calculus 10e by Larson and Edwards, p. 1034. Exercises 3, 9, 13, 15, 17.

Exercise 3 Evaluating an Iterated Integral

Evaluate the iterated integral.

$$\int_{0}^{1} \int_{0}^{1+x} (3x+2y) \: dy \: dx$$

Solution

$$\int_{0}^{1} \int_{0}^{1+x} (3x+2y) \: dy \: dx$$ $$= \int_{0}^{1} \int_{0}^{1+x} 3x \: dy+ \int_{0}^{1+x} 2y \: dy \: dx $$ Apply the Sum Rule
$$= \int_{0}^{1} \left. \vphantom{\frac{3}{4}} 3xy \right]_{0}^{1+x} + \left. \vphantom{\frac{3}{4}} y^{2} \right]_{0}^{1+x} \: dx $$
$$= \int_{0}^{1} (x+1)(4x+1) \: dx$$
$$= \left. \vphantom{\frac{3}{4}} 2x^{3}- \frac{2}{3} x^{3} +2x^{2} + \frac{1}{2} x^{2} \right]_{0}^{1} = \frac{29}{6}$$

Exercise 9 Finding the Area for a Region

Use an iterated integral to find the area for the region bounded by the graphs.

\(y=x, \: y=2x+2, \: x=0, \: x=4\)

Solution

Figure 1

The graph is shown in Figure 1. The equation for the area is

$$\int_{0}^{4} \int_{x}^{2x+2} \: dy \: dx $$

Evaluating the integral produces

$$\int_{0}^{4} \int_{x}^{2x+2} \: dy \: dx $$ $$= \left. \int_{0}^{4} y \right]_{x}^{2x+2} \: dx $$
$$= \int_{0}^{4} x+2 \: dx $$
$$= \left. \frac{1}{2}x^{2} + 2x \right]_{0}^{4}$$
$$= \frac{1}{2}(4)^{2} + 2(4) $$
$$= \frac{1}{2}(16) + 8 = 16$$

Exercise 13 Switching the Integration Order

Sketch the region \(R\) whose area is given by the iterated integral. Then switch the integration order and show that both orders yield the same area.

$$ \int_{0}^{4} \int_{2x}^{8} \: dy \: dx$$

Solution

Figure 2

The sketch is shown in Figure 2. The first integration order, \(dy \: dx\), produces

$$ \int_{0}^{4} \int_{2x}^{8} \: dy \: dx $$ $$= \left. \int_{0}^{4} y \right]_{2x}^{8} \: dx $$
$$= \int_{0}^{4} 8-2x \: dx$$
$$= \left. \vphantom{\frac{1}{2}} 8x-x^{2} \right]_{0}^{4}$$
$$=8(4)-(4)^{2}] = 16$$

The second integration order, \(dx \: dy\), requires new bounds, \(0 \leqslant x \leqslant 1/2y\) and \(0 \leqslant y \leqslant 8 \), which produces

$$ \int_{0}^{8} \int_{0}^{1/2y} \: dx \: dy $$ $$= \left. \int_{0}^{8} x \right]_{0}^{1/2y} \: dy $$
$$= \int_{0}^{8} \frac{1}{2}y \: dy $$
$$= \left. \frac{1}{4}y^{2} \right]_{0}^{8} = 16$$

The areas are equal.

Exercise 15 Evaluating a Double Integral

Set up integrals for both integration orders. Use the more convenient order to evaluate the integral over the region \(R\).

$$ \int_{R} \int 4xy \: dA$$
\(R\): rectangle with vertices (0,0), (0,4), (2,4), (2,0).
Figure 3

Solution The graph is shown in Figure 3. The bounds are \(0 \leqslant x \leqslant 2\) and \(0 \leqslant y \leqslant 4\). The integral for the area are

$$ \int_{0}^{2} \int_{0}^{4} 4xy \: dy \: dx \text{ and } \int_{0}^{4} \int_{0}^{2} 4xy \: dx \: dy. $$

Since both are equally easy to solve the first will be used.

$$\int_{0}^{2} \int_{0}^{4} 4xy \: dy \: dx$$ $$=4 \int_{0}^{2} \left. \frac{1}{2} xy^{2} \right]_{0}^{4} \: dx$$
$$= 2 \int_{0}^{2} 16x \:dx $$
$$= 32 \left[ \frac{1}{2} x^{2} \right]_{0}^{2}= 64 $$

Exercise 17 Finding Volume

Figure 4

Use a double integral to find the volume for the solid in Figure 4.
Solution The bounds are \(0 \leqslant x \leqslant 3\) and \(0 \leqslant y \leqslant 2 \). The equation for the volume is.

$$ Volume$$ $$=\int_{0}^{3} \int_{0}^{2} (5-x) \: dy \: dx $$
$$= \left. \int_{0}^{3} y(5-x) \right]_{0}^{2} \: dx $$
$$= \int_{0}^{3} 10-2x \: dx $$
$$= \left. \vphantom{\frac{1}{2}} 10x-x^{2} \right]_{0}^{3} $$
$$= 10(3)-(3)^{3} = 21$$

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Parent Article: Calculus III Advanced (Course)