Calculus III Advanced (Course) (14.06) (Homework)

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Section 14.6 Homework

From Calculus 10e by Larson and Edwards, p. 1017. Exercises 6, 14, 21, 26, 36, 38

Exercise 14.6.6 Evaluating a Triple Iterated Integral

Evaluate the triple iterated integral.

$$ \int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1/xz} \ln z \: dy \: dz \: dx $$

Solution

$$ \int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1/xz} \ln z \: dy \: dz \: dx$$ $$ =\left. \int_{1}^{4} \int_{1}^{e^{2}} y\ln z \right]_{0}^{1/xz} \: dz \: dx $$
$$= \int_{1}^{4} \int_{1}^{e^{2}} \frac{1}{xz}\ln z \: dz \: dx$$
$$= \int_{1}^{4} \left. \frac{1}{2x} \ln^{2}(z) \right]_{1}^{e^{2}}\: dx$$
$$= \int_{1}^{4} \frac{1}{2x} \ln^{2}(e^{2}) - \frac{1}{2x} \ln^{2}(1) \: dx $$
$$= \int_{1}^{4} \frac{1}{2x}\left[ \ln^{2}(e^{2}) - \ln^{2}(1) \right] \: dx $$
$$= \int_{1}^{4} \frac{2}{x} \: dx $$
$$= \left. \vphantom{ \frac{1}{2}} 2 \ln(x) \right]_{1}^{4} $$
$$= 2 [\ln(4) - \ln(1)] = 4 \ln(2) $$

Exercise 14.6.14 Setting Up a Triple Integral

Set up a triple integral for the solid bounded by

\( z= \sqrt{16-x^{2}-y^{2}}\)

and

\(z=0.\)

Solution The solid is a hemisphere positive to the \(xy\)-plane, as shown in Figure 1.

Figure 1

The bounds are

\(-4 \leqslant x \leqslant 4,\)
\(-4 \leqslant y \leqslant 4,\)

and

\(0 \leqslant z \leqslant \sqrt{16-x^{2}-y^{2}}.\)

This produces the triple integral for the volume

$$V $$ $$= \int \int_{Q} \int \: dV$$
$$=\int_{-4}^{4} \int_{-4}^{4} \int_{0}^{\sqrt{16-x^{2}-y^{2}}} \: dz \:dx \:dy $$

Exercise 14.6.21 Volume

Use a triple integral to find the volume for the solid bounded by the graphs for:

\(z=4-x^{2}, \)
\(y=4-x^{2}, \)

and the first octant.
Solution The bounds are:

\( 0 \leqslant z \leqslant 4-x^{2}, \)
\( 0 \leqslant y \leqslant 4-x^{2}, \)

and

\( 0 \leqslant x \leqslant 2.\)

This produces

$$ \int \int \int \: dV $$ $$= \int_{0}^{2} \int_{0}^{4-x^{2}} \int_{0}^{4-x^{2}} \: dz \: dy \: dx$$
$$= \int_{0}^{2} \int_{0}^{4-x^{2}} (4-x^{2}) \: dy \: dx$$
$$= \left. \int_{0}^{2} y(4-x^{2}) \right]_{0}^{4-x^{2}} \: dx $$
$$= \int_{0}^{2} (4-x^{2})^{2} \: dx$$
$$= \int_{0}^{2} (16-8x^{2}+x^{4}) \:dx $$
$$= \left[16-\frac{8}{3}x^{3}+\frac{1}{5}x^{5} \right]_{0}^{2} = \frac{256}{15}$$

Exercise 14.6.26 Change the Integration Order

Sketch the solid and rewrite the integral using the indicated integration order.

$$ \int_{-1}^{1} \int_{y^{2}}^{1} \int_{0}^{1-x} \: dz \: dx \: dy $$

Rewrite using the order \(dx \: dz \: dy\).
Solution The bounds are

\( y^{2} \leqslant x \leqslant 1,\)
\( -1 \leqslant y \leqslant 1,\)

and

\( 0 \leqslant z \leqslant 1-x.\)

The projection on the \(xy\)-plane is the region:

\( 0 \leqslant z \leqslant 1-y^{2}, \: -1 \leqslant y \leqslant 1.\)

Where \(y^{2} = x\).

Figure 2

$$ \int_{-1}^{1} \int_{0}^{1-x} \int_{y^{2}}^{1} \: dx \: dz \: dy $$

Exercise 14.6.36 Integration Orders

Figure 3 shows the integration region for the integral. Rewrite the integral as an equivalent iterated integral in the five other orders.

$$ \int_{0}^{3} \int_{0}^{x} \int_{0}^{9-x^{2}} \: dz \: dy \: dx $$
Figure 3

Solution

$$\textbf{1. } \int_{0}^{9-x^{2}} \int_{0}^{3} \int_{0}^{x} \: dy \: dz \: dx $$
$$\textbf{2. } \int_{0}^{3} \int_{0}^{9-x^{2}} \int_{0}^{x} \: dy \: dx \: dz $$
$$\textbf{3. } \int_{0}^{3} \int_{0}^{x} \int_{0}^{9-x^{2}} \: dx \: dy \: dz $$
$$\textbf{4. } \int_{0}^{x} \int_{0}^{9-x^{2}} \int_{0}^{3} \: dz \: dx \: dy $$
$$\textbf{5. } \int_{0}^{x} \int_{0}^{3} \int_{0}^{9-x^{2}} \: dx \: dz \: dy $$

Exercise 14.6.38 Mass and Mass Center

Find the mass and the indicated coordinates for the mass center for the solid region \(Q\) with density \(\rho\) bounded by the graphs.
Find \(\bar{y}\) using \(\rho(x,y,z)= ky\).

\(Q: \: 3x+3y+5z=15, \: x=0, \: y=0, \: z=0\)

Solution

$$ \int \int \int_{Q} \rho(x,y,z) \: dV$$ $$= \int_{0}^{5} \int_{0}^{5-y} \int_{0}^{(15-3x-3y)/5} ky \: dz \: dx \: dy $$
$$= \frac{1}{5} k \int_{0}^{5} \int_{0}^{5-y} y (15-3x-3y) \: dx \: dy $$
$$= \frac{1}{5} k \int_{0}^{5} \int_{0}^{5-y} 15y-3xy-3y^{2} \: dx \: dy $$
$$= \frac{1}{5} k \int_{0}^{5} \left[ 3y(5-y)x-3y \frac{1}{2} x^{2} \right]_{0}^{5-y} \: dy $$
$$= \frac{1}{5} k \int_{0}^{5} \left( 3y(5-y)^{2}-3y \frac{1}{2} (5-y)^{2} \right) \: dy $$
$$= \frac{3k}{10} \int_{0}^{5} y(5-y)^{2} \: dy $$
$$= \frac{3k}{10} \int_{0}^{5} (25y-10y^{2}+y^{3}) \: dy $$
$$= \frac{3k}{10} \left[ \frac{25}{2}y^{2} - \frac{10}{3} y^{3} + \frac{1}{4} y^{4} \right]_{0}^{5}$$
$$= \frac{3k}{10} (25)^{2} \left( \frac{1}{2} - \frac{2}{3} + \frac{1}{4} \right) = \frac{125k}{8}$$

The integral for the \(y\) coordinate is:

$$ \int \int \int_{Q} y \rho(x,y,z) \: dV $$ $$= \int \int_{R_{xy}} \int_{0}^{(15-3x-3y)/5} yky \: dz \: dA$$
$$= \frac{1}{5} k \int_{0}^{5} \left[ 3y^{2}(5-y)x - 3y \frac{1}{2} x^{2} \right]_{0}%{5-y}$$
$$= \frac{3k}{10} \int_{0}^{5} y^{2}(5-y)^{2} \: dy$$
$$= \frac{3k}{10} \int_{0}^{5} (25y^{2} -10y^{3} +y^{4}) \: dy $$
$$= \frac{3k}{10} \left[ 25 \frac{y^{3}}{3} - \frac{5}{2} y^{4} + \frac{1}{5} y^{5} \right]_{0}^{5} = \frac{125k}{4} $$
$$ \bar{y} = \frac{\frac{125k}{4}}{\frac{125k}{8}} = 2$$

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Parent Article: Calculus III Advanced (Course)