Bài tập 3.47 trang 181 SBT Toán 12

Bài tập 3.47 trang 181 SBT Toán 12

a) Ta có: \(y' = \frac{2}{3}{x^{ - \frac{1}{3}}}\)

Với x = 1 thì y = 1 và \(y'\left( 1 \right) = \frac{2}{3}\).

Tiếp tuyến \(y = \frac{2}{3}\left( {x - 1} \right) + 1 = \frac{2}{3}x + \frac{1}{3}\)

Có \(y = {x^{\frac{2}{3}}} \Rightarrow x = {y^{\frac{3}{2}}}\) và

\(y = \frac{2}{3}x + \frac{1}{3} \Rightarrow x = \frac{3}{2}y - \frac{1}{2}\)

Khi đó \({y^{\frac{3}{2}}} = \frac{3}{2}y - \frac{1}{2} \Rightarrow y = 1\).

Ta có: \(\frac{3}{2}y - \frac{1}{2} = 0 \Leftrightarrow y = \frac{1}{3}\)

\(\begin{array}{l}
V = \pi \int \limits_0^1 {\left( {{y^{\frac{3}{2}}}} \right)^2}dy - \pi \int \limits_{\frac{1}{3}}^1 {\left( {\frac{3}{2}y - \frac{1}{2}} \right)^2}dy\\
 = \pi \int \limits_0^1 {y^3}dy - \pi \int \limits_{\frac{1}{3}}^1 {\left( {\frac{3}{2}y - \frac{1}{2}} \right)^2}dy\\
 = \pi .\left. {\frac{{{y^4}}}{4}} \right|_0^1 - \pi \int\limits_{\frac{1}{3}}^1 {\left( {\frac{9}{4}{y^2} - \frac{3}{2}y + \frac{1}{4}} \right)dy} \\
 = \frac{\pi }{4} - \pi .\left. {\left( {\frac{3}{4}{y^3} - \frac{3}{4}{y^3} + \frac{1}{4}y} \right)} \right|_{\frac{1}{3}}^1\\
 = \frac{\pi }{4} - \frac{{2\pi }}{9} = \frac{\pi }{{36}}
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
\frac{1}{x} - 1 = 2x \Rightarrow x = \frac{1}{2}\\
\frac{1}{x} - 1 = 0 \Leftrightarrow x = 1\\
2x = 0 \Leftrightarrow x = 0
\end{array}\)

Do đó 

\(\begin{array}{l}
V = \pi \int \limits_0^{\frac{1}{2}} {\left( {2x} \right)^2}dx + \pi \int \limits_{\frac{1}{2}}^1 {\left( {\frac{1}{x} - 1} \right)^2}dx\\
 = \pi .\int \limits_0^{\frac{1}{2}} 4{x^2}dx + \pi .\int \limits_{\frac{1}{2}}^1 \left( {\frac{1}{{{x^2}}} - \frac{2}{x} + 1} \right)dx\\
 = \pi .\left. {\frac{{4{x^3}}}{3}} \right|_0^{\frac{1}{2}} + \pi \left. {\left( { - \frac{1}{x} - 2\ln x + x} \right)} \right|_{\frac{1}{2}}^1\\
 = \frac{\pi }{6} + \pi \left( {0 + 2 + 2\ln \frac{1}{2} - \frac{1}{2}} \right)\\
 = \frac{\pi }{6} + \frac{{3\pi }}{2} - 2\pi \ln 2\\
 = \frac{{5\pi }}{3} - 2\pi \ln 2
\end{array}\)

c) +) Quay quanh Ox${\displaystyle Ox}$.

Ta có: \(\left| {2x - {x^2}} \right| = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2.
\end{array} \right.\)

Khi đó

\(\begin{array}{l}
V = \pi \int \limits_0^3 {\left( {2x - {x^2}} \right)^2}dx\\
 = \pi \int \limits_0^3 \left( {4{x^2} - 4{x^3} + {x^4}} \right)dx\\
 = \pi \left. {\left( {\frac{{4{x^3}}}{3} - {x^4} + \frac{{{x^5}}}{5}} \right)} \right|_0^3\\
 = \pi \left( {\frac{{4.27}}{3} - {3^4} + \frac{{{3^5}}}{5}} \right) = \frac{{18\pi }}{5}
\end{array}\)

+) Quay quanh Oy.

Ta có: 

\(\begin{array}{l}
y = \left| {2x - {x^2}} \right| \Rightarrow \left[ \begin{array}{l}
y = 2x - {x^2}\\
y =  - 2x + {x^2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x + y = 0\\
{x^2} - 2x - y = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1 \pm \sqrt {1 - y} \\
x = 1 \pm \sqrt {1 + y} 
\end{array} \right.
\end{array}\)

Ta có: 

\(\begin{array}{l}
{V_y} = \pi \int \limits_0^1 \left[ {{{\left( {1 + \sqrt {1 - y} } \right)}^2} - {{\left( {1 - \sqrt {1 - y} } \right)}^2}} \right]dy\\
 + \pi \int \limits_0^3 \left[ {{3^2} - {{\left( {1 + \sqrt {1 + y} } \right)}^2}} \right]dy\\
 = \pi \int \limits_0^1 \left( {1 + 2\sqrt {1 - y}  + 1 - y - 1 + 2\sqrt {1 - y}  - 1 + y} \right)dy\\
 + \pi \int\limits_0^3 {\left( {9 - 1 - 2\sqrt {1 + y}  - 1 - y} \right)dy} \\
 = \pi \int \limits_0^1 4\sqrt {1 - y} dy + \pi \int \limits_0^3 \left( {7 - y - 2\sqrt {1 + y} } \right)dy\\
 = 4\pi \int \limits_0^1 \sqrt {1 - y} dy + \pi \left[ {\left. {\left( {7y - \frac{{{y^2}}}{2}} \right)} \right|_0^3 - 2\int\limits_0^3 {\sqrt {1 + y} dy} } \right]\\
 = 4\pi I + \pi (\frac{{33}}{2} - 2J)
\end{array}\)

Tính \(I = \int \limits_0^1 \sqrt {1 - y} dy\) ta có:

Đặt \(\sqrt {1 - y}  = t \Rightarrow 1 - y = {t^2}\)

\(\begin{array}{l}
 \Rightarrow  - dy = 2tdt \Rightarrow dy =  - 2tdt\\
 \Rightarrow I = \int \limits_1^0 t.\left( { - 2tdt} \right) = \int\limits_0^1 {2{t^2}dt} \\
 = \left. {\frac{2}{3}{t^3}} \right|_0^1 = \frac{2}{3}
\end{array}\)

Tính \(J = \int \limits_0^3 \sqrt {1 + y} dy\) ta có:

Đặt \(t = \sqrt {1 + y}  \Rightarrow {t^2} = 1 + y\)

\(\Rightarrow 2tdt = dy\)

\( \Rightarrow J = \int\limits_1^2 {t.2tdt}  = \left. {\frac{{2{t^3}}}{3}} \right|_1^2 = \frac{{14}}{3}\)

Vậy \(V = 4\pi .\frac{2}{3} + \pi \left( {\frac{{33}}{2} - 2.\frac{{14}}{3}} \right) = \frac{{59\pi }}{6}\).

-- Mod Toán 12