Volume Benda Putar Terhadap Sumbu-x
Volume benda putar jika daerah yang dibatasi kurva $y = f(x)$, sumbu-x, garis $(x = a)$ dan $(x = b)$ diputar $360^o$ mengelilingi sumbu-x adalah :
$V=\pi\int_{a}^{b}y^{2}\:dx \;\;atau \;\;V=\pi\int_{a}^{b}\left [ f(x) \right ]^{2}\:dx$
Volume benda putar jika daerah yang dibatasi kurva $y_1 = f(x), y_2 = g(x)$, garis $(x = a)$ dan $(x = b)$ diputar $360^o$ mengelilingi sumbu-x adalah : $V=\pi\int_{a}^{b}\left (y{_{1}}^{2}-y{_{2}}^{2} \right )\:dx \;\;atau \;\;V=\pi\int_{a}^{b}\left ( \left [f(x) \right ]^{2}-\left [g(x) \right ]^{2} \right )\:dx$
Volume Benda Putar Terhadap Sumbu-y
Volume benda putar jika daerah yang dibatasi kurva $x = f(y)$, sumbu-y, garis $(y = a)$ dan $(y = b)$ diputar $360^o$ mengelilingi sumbu-y adalah : $V=\pi\int_{a}^{b}x^{2}\:dy\;\;atau\;\; V=\pi\int_{a}^{b}\left [ f(y) \right ]^{2}\:dy$
Volume benda putar jika daerah yang dibatasi kurva $x_1 = f(y), x_2 = g(y)$, garis $(x = a)$ dan $(x = b)$ diputar $360^o$ mengelilingi sumbu-y adalah : $V=\pi\int_{a}^{b}\left (x{_{1}}^{2}-x{_{2}}^{2} \right )\:dy\;\;atau\;\;V=\pi\int_{a}^{b}\left ( \left [f(y) \right ]^{2}-\left [g(y) \right ]^{2} \right )\:dy$
Contoh 1
Volume Benda putar yang terbentuk jika daerah yang dibatasi kurva $y=2x-x^{2},\;\; sumbu-x,\;\; 0\leq x\leq 1$, diputar $360^o$ mengelilingi sumbu-x adalah... satuan volume.
Jawab :
Titik potong sumbu-x ⇒ y = 0
2x − $x^2$ = 0
x(2 − x) = 0
x = 0 atau x = 2
$V = \pi \int_{0}^{1}y^2 dx$
$V = \pi \int_{0}^{1}(2x-x^2)2 dx$
$V = \pi \int_{0}^{1}(x^4 -4x^3 + 4x^2) dx$
$V = \pi \left [\frac{1}{5}x^{5}-x^{4}+\frac{4}{3}x^{3} \right ]_{0}^{1}$
$V = \frac{8}{15}\pi $
Contoh 2
Volume benda putar yang terjadi jika daerah diantara kurva $y=\sqrt{x}\;\;\ dan\;\;y=\frac{1}{2}x$, diputar $360^o$ mengelilingi sumbu-x adalah... satuan volume.
Jawab :
Misalkan :
$y_1 = \sqrt{x}$
$y_2 = \frac{1}{2}x$
Titik potong kurva :
$y_1 = y_2$
$\sqrt{x} = \frac{1}{2}x\;\; (kuadratkan)$
$x = \frac{1}{4}x^2\;\; (kali\;4)$
$4x = x^2$
$4x - x^2 = 0$
$x (4 - x) = 0$
x = 0 atau x = 4
$V =\pi \int_{0}^{4}(y_1^2-y_2^2)\;\; dx$
$V =\pi \int_{0}^{4} \left (\sqrt{x} \right )^{2}-\left (\frac{1}{2}x \right )^{2}\;dx$
$V =\pi \int_{0}^{4}(x -\frac{1}{4}x)^2\; dx$
$V =\pi \left [ \frac{1}{2}x^{2}-\frac{1}{12}x^{3} \right ]_{0}^{4}$
$V = \frac{8}{3}\pi $
$V =\pi \int_{0}^{4} \left (\sqrt{x} \right )^{2}-\left (\frac{1}{2}x \right )^{2}\;dx$
$V =\pi \int_{0}^{4}(x -\frac{1}{4}x)^2\; dx$
$V =\pi \left [ \frac{1}{2}x^{2}-\frac{1}{12}x^{3} \right ]_{0}^{4}$
$V = \frac{8}{3}\pi $
Contoh 3
Daerah yang dibatasi kurva $y=x^{2}$, garis $y=2-x$ dan sumbu-x diputar diputar $360^o$ mengelilingi sumbu-x. Volume benda putar yang terjadi adalah ... satuan volume.
Jawab :
Misalkan :
$y_1 = x^2$
$y_2 = 2- x$
Titik potong kurva :
$y_1 = y_2$
$x^2= 2- x$
$x^2 + x-2 = 0$
(x + 2)(x -1) = 0$
x = −2 atau x = 1
Titik potong garis dan sumbu-x ⇒ y = 0
2 − x = 0
x = 2
$V_I =\pi \int_{0}^{1}y_1^2\; dx$
$V_I =\pi \int_{0}^{1}\; (x^2)^2\; dx$
$V_I =\pi \int_{0}^{1}\; x^4 \;dx$
$V_I = \pi \left [\frac{1}{5}x^{5} \right ]_{0}^{1}$
$V_I = \frac{1}{5}\pi $
$V_I =\pi \int_{0}^{1}\; x^4 \;dx$
$V_I = \pi \left [\frac{1}{5}x^{5} \right ]_{0}^{1}$
$V_I = \frac{1}{5}\pi $
$V_{II} = \pi \int_{1}^{2}y_2^2\; dx$
$V_{II} = \pi \int_{1}^{2} (2-x)^2 \;dx$
$V_{II} =\pi \int_{1}^{2} (x^2-4x + 4)\;\; dx$
$V_{II} =\pi \left [ \frac{1}{3}x^{3}-2x^{2}+4x \right ]_{1}^{2}$
$V_{II} =\frac{1}{3}\pi $
$V_{II} = \pi \int_{1}^{2} (2-x)^2 \;dx$
$V_{II} =\pi \int_{1}^{2} (x^2-4x + 4)\;\; dx$
$V_{II} =\pi \left [ \frac{1}{3}x^{3}-2x^{2}+4x \right ]_{1}^{2}$
$V_{II} =\frac{1}{3}\pi $
Sehingga diperoleh :
$V = V_{I} + V_{II}$
$V = \frac{1}{5}\pi + \frac{1}{3}\pi $
$V = \frac{8}{15}\pi $
Contoh 4
Volume benda putar yang terjadi jika daerah yang dibatasi kurva $y^{2}=2x+4$ dan sumbu-y dikuadran kedua, diputar $360^o$ mengelilingi sumbu-y adalah ... satuan volume.
Jawab :
$y^2 = 2x + 4$
⇒ $2x = y^2- 4$
⇒ $x = \frac{1}{2}y^2-2$
Titik potong kurva dan sumbu-y ⇒ x = 0
$\frac{1}{2}y^2-2 = 0\; (kali\; 2)$
$y^2 - 4 = 0$
$(y + 2)(y- 2) = 0$
y = −2 atau y = 2
$V = \pi \int_{0}^{2} x^2 \;\;dy$
$V = \pi \int_{0}^{2} ((\frac{1}{2})y^2-2)^2\;\; dy$
$V = \pi \int_{0}^{2} ((\frac{1}{4})y^4-2y^2 + 4)\;\; dy$
$V = \pi \left [ \frac{1}{20}y^{5}-\frac{2}{3}y^{3}+4y \right ]_{0}^{2}$
$V =\frac{64}{15}\pi $
$V = \pi \int_{0}^{2} ((\frac{1}{2})y^2-2)^2\;\; dy$
$V = \pi \int_{0}^{2} ((\frac{1}{4})y^4-2y^2 + 4)\;\; dy$
$V = \pi \left [ \frac{1}{20}y^{5}-\frac{2}{3}y^{3}+4y \right ]_{0}^{2}$
$V =\frac{64}{15}\pi $
Contoh 5
Volume benda putar yang terbentuk bila daerah antara kurva $y=x^{2}-4$ dan $y=2x-4$ diputar 360o mengelilingi sumbu-y adalah ... satuan volume.
Jawab :
$y = x^2 -4$
⇒ $x^2 = y + 4$
$y = 2x -4$
⇒ $2x = y + 4$
⇒ $x =\frac{1}{2}y + 2$
⇒ $x^2 = (\frac{1}{2}y + 2)^2$
Misalkan :
$x_1^2 = y + 4$
$x_2^2 = (\frac{1}{2}y + 2)^2$
Titik potong kurva :
$x_1^2 = x_2^2$
$y + 4 = (\frac{1}{2}y + 2)^2$
$y + 4 = \frac{1}{4}y^2 + 2y + 4$
$\frac{1}{4}y^2 + y = 0 (kali 4)$
$y^2 + 4y = 0$
$y(y + 4) = 0$
$y + 4 = (\frac{1}{2}y + 2)^2$
$y + 4 = \frac{1}{4}y^2 + 2y + 4$
$\frac{1}{4}y^2 + y = 0 (kali 4)$
$y^2 + 4y = 0$
$y(y + 4) = 0$
y = 0 atau y = −4
$V = \pi \int_{-4}^{0}(x_1^2- x_2^2) \;\;dx$
$V = \pi \int_{-4}^{0}{(y + 4)- ((\frac{1}{4})y^2 + 2y + 4)} \;\;dx$
$V = \pi \int_{-4}^{0}((-\frac{1}{4})y^2-y)\;\; dx$
$V = \pi \left [-\frac{1}{12}y^{3}-\frac{1}{2}y^{2} \right ]_{-4}^{0}$
$V = \frac{8}{3}\pi $
$V = \pi \int_{-4}^{0}{(y + 4)- ((\frac{1}{4})y^2 + 2y + 4)} \;\;dx$
$V = \pi \int_{-4}^{0}((-\frac{1}{4})y^2-y)\;\; dx$
$V = \pi \left [-\frac{1}{12}y^{3}-\frac{1}{2}y^{2} \right ]_{-4}^{0}$
$V = \frac{8}{3}\pi $
![Pembahasan Soal Matematika - Barisan Aritmatika [versi 1]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMtba1SEosevhJ1Igskc9gjRb-0S_CXsJcgP2iIOJVg7NohrjUretEAAiSJvEGU29q3rWn93M5w-8-FPjWRlHhKIZ_p299jXAxxmxVy1Vyvs0spFaqT0_0mrtdm6p0Ee5tE8Bp9BCsjBuV/s220/Barisan+Aritmatika+soal.png "Pembahasan Soal Matematika - Barisan Aritmatika [versi 1]")
