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Chapter 5: Q.2 (page 432)

Another way of thinking of the volume in question is as an accumulation of nested open cylinders, or 鈥渟hells,鈥� as y varies from 0 to ln 3, as shown previously at the right. The shell at a given y 鈭� [0, ln 3] has a radius r = y and length $h = 3 鈭� f^{鈭� 1} (y)$ , and thus the surface area $2 蟺 y (3 - f^{- 1} (y))$ , where $f^{鈭� 1} (y) = e^{y}$ here is the inverse function of f (x) = ln x. As we will see in Section 6.2, the definite integral,
$2 蟺 {鈭�}_{0}^{\ln 3} y (3 - f^{- 1} (y)) d y$

Short Answer

Expert verified

The result is $2 蟺 {鈭�}_{0}^{\ln 3} y (3 - e^{y}) d y$ .

Step by step solution

01

Step 1. Given information.

Consider the given information.

$2 蟺 {鈭�}_{0}^{\ln 3} y (3 - f^{- 1} (y)) d y$ and $y = e^{y}$

02

Step 2. Calculation.

Substitute the function in the integral and calculate the integral.

$\begin{array}{rcl} 2 蟺 {鈭�}_{0}^{\ln 3} y (3 - f^{- 1} (y)) d y & = & 2 蟺 {鈭�}_{0}^{\ln 3} y (3 - e^{y}) d y \\ = & 2 蟺 {鈭�}_{0}^{\ln 3} (3 y - y e^{y}) d y \\ = & 2 蟺 {[\frac{3 y^{2}}{2} - 鈭� y e^{y} d y]}_{0}^{\ln 3} \\ = & 2 蟺 {[\frac{3 y^{2}}{2} - [y e^{y} - 鈭� e^{y} d y]]}_{0}^{\ln 3} \\ = & 2 蟺 {[\frac{3 y^{2}}{2} - y e^{y} + e^{y}]}_{0}^{\ln 3} \\ = & 2 蟺 [(\frac{3 {(\ln 3)}^{2}}{2} - \ln 3 e^{\ln 3} + e^{\ln 3}) - (\frac{3 {(0)}^{2}}{2} - 0 e^{0} + e^{0})] \\ = & 2 蟺 [\frac{3 {(\ln 3)}^{2}}{2} - 3 \ln 3 + 3 - 1] \\ = & 2 蟺 [\frac{3 {(\ln 3)}^{2}}{2} - 3 \ln 3 + 2] \end{array}$

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Most popular questions from this chapter

Solve given definite integral.

${鈭�}_{1}^{2} 鈥� \frac{3}{{(9 鈭� 2 x^{2})}^{3 / 2}} d x$

Describe two ways in which the long-division algorithm for polynomials is similar to the long-division algorithm for integers and then two ways in which the two algorithms are different.

Solve the integral $鈭� x^{3} \sqrt{x^{2} - 1} d x$ three ways:

(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;

(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$

(c) with the trigonometric substitution x = sec u.

Solve given definite integral.

${鈭�}_{0}^{4} 鈥� x^{3} \sqrt{x^{2} + 4} d x$

Write $\sin (2 \cos^{- 1} x)$ as an algebraic function.

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