Exercise Zone : Volume Benda Putar

Berikut ini adalah kumpulan soal mengenai Volume Benda Putar. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

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No. 1

Volume benda putar yang terjadi jika daerah yang dibatasi kurva y=x^2-4, sumbu X, garis x=0 dan x=1 diputar mengelilingi sumbu X adalah ....

1. \dfrac{103}{15}\pi satuan volume
2. \dfrac{166}{15}\pi satuan volume
3. \dfrac{203}{15}\pi satuan volume

4. \dfrac{211}{15}\pi satuan volume
5. \dfrac{243}{15}\pi satuan volume

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$$\begin{array}{rl}V& =\pi {\displaystyle \underset{0}{\overset{1}{\int }}{(x^2-4)}^{2}dx}\\ & =\pi {\displaystyle \underset{0}{\overset{1}{\int }}(x^4-8x^2+16)dx}\\ & ={[{\displaystyle \frac{1}{5}}{x}^5-{\displaystyle \frac{8}{3}}{x}^3+16x]}_0^1\\ & =[{\displaystyle \frac{1}{5}}(1{)}^5-{\displaystyle \frac{8}{3}}(1{)}^3+16(1)]-[{\displaystyle \frac{1}{5}}(0{)}^5-{\displaystyle \frac{8}{3}}(0{)}^3+16(0)]\\ & =\pi ({\displaystyle \frac{1}{5}}-{\displaystyle \frac{8}{3}}+16)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{203}{15}}\pi }}}}\end{array}$$

Jadi, Volume benda putar yang terjadi adalah \dfrac{203}{15}\pi. JAWAB: C

No. 2

Daerah A dibatasi oleh grafik kurva y=x\sin x dengan x\geq0, sumbu x, garis x=0 dan garis x=\pi. Volume benda putar yang terjadi jika daerah A diputar mengelilingi sumbu x adalah sebesar \dfrac{\pi^2}6\left(\pi^2-a\right), maka nilai a=

SUMBER

$$\begin{array}{rl}V& =\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}(x\sin x{)}^{2}dx}\\ & =\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}{x}^2{\sin }^{2}xdx}\\ & =\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}{x}^2({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}\cos 2x)dx}\\ & =\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{1}{2}}{x}^2-{\displaystyle \frac{1}{2}}{x}^2\cos 2xdx}\end{array}$$

u	dv
+-\dfrac12x^2	\cos2x
--x	\dfrac12\sin2x
+-1	-\dfrac14\cos2x
-0	-\dfrac18\sin2x

$$\begin{array}{rl}V& =\pi {[{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{3}}{x}^3+(-{\displaystyle \frac{1}{2}}{x}^2)({\displaystyle \frac{1}{2}}\sin 2x)-(-x)(-{\displaystyle \frac{1}{4}}\cos 2x)+(-1)(-{\displaystyle \frac{1}{8}}\sin 2x)]}_0^{\pi }\\ & =\pi {[{\displaystyle \frac{1}{6}}{x}^3-{\displaystyle \frac{1}{4}}{x}^2\sin 2x-{\displaystyle \frac{1}{4}}x\cos 2x+{\displaystyle \frac{1}{8}}\sin 2x]}_0^{\pi }\\ & =\pi [({\displaystyle \frac{1}{6}}{\pi }^3-{\displaystyle \frac{1}{4}}{\pi }^2\sin 2\pi -{\displaystyle \frac{1}{4}}\pi \cos 2\pi +{\displaystyle \frac{1}{8}}\sin 2\pi )-({\displaystyle \frac{1}{6}}(0{)}^3-{\displaystyle \frac{1}{4}}(0{)}^2\sin 2(0)-{\displaystyle \frac{1}{4}}(0)\cos 2(0)+{\displaystyle \frac{1}{8}}\sin 2(0))]\\ & =\pi [({\displaystyle \frac{1}{6}}{\pi }^3-{\displaystyle \frac{1}{4}}{\pi }^2(0)-{\displaystyle \frac{1}{4}}\pi (1)+{\displaystyle \frac{1}{8}}(0))-0]\\ & =\pi ({\displaystyle \frac{1}{6}}{\pi }^3-{\displaystyle \frac{1}{4}}\pi )\\ & ={\displaystyle \frac{{\pi }^2}{6}}({\pi }^2-{\displaystyle \frac{6}{4}})\\ & ={\displaystyle \frac{{\pi }^2}{6}}({\pi }^2-{\displaystyle \frac{3}{2}})\end{array}$$

Jadi, a=\dfrac32

No. 3

Tentukanlah luas daerah bidang berikut dan tentukan pula volumenya seandainya bidang yang diarsir tersebut diputar terhadap sumbu X

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$$\begin{array}{rl}L& ={\displaystyle \underset{1}{\overset{3}{\int }}2xdx}\\ & ={x^2|}_1^3\\ & =3^2-1^2\\ & =9-1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$$

$$\begin{array}{rl}V& =\pi {\displaystyle \underset{1}{\overset{3}{\int }}(2x{)}^{2}dx}\\ & =\pi {\displaystyle \underset{1}{\overset{3}{\int }}4x^{2}dx}\\ & =\pi {\left[{\displaystyle \frac{4}{3}}{x}^3\right]}_1^3\\ & =\pi [{\displaystyle \frac{4}{3}}(3{)}^3-{\displaystyle \frac{4}{3}}(1{)}^3]\\ & =\pi [{\displaystyle \frac{4}{3}}(27)-{\displaystyle \frac{4}{3}}(1)]\\ & =\pi [36-{\displaystyle \frac{4}{3}}]\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 34{\displaystyle \frac{2}{3}}\pi }}}}\end{array}$$

Jadi, L=8, V=34\dfrac23\pi