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

---

No.

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

Facebook

\begin{aligned} V&=\pi\displaystyle\intop_0^1\left(x^2-4\right)^2\ dx\\ &=\pi\displaystyle\intop_0^1\left(x^4-8x^2+16\right)\ dx\\ &=\left[\dfrac15x^5-\dfrac83x^3+16x\right]_0^1\\ &=\left[\dfrac15(1)^5-\dfrac83(1)^3+16(1)\right]-\left[\dfrac15(0)^5-\dfrac83(0)^3+16(0)\right]\\ &=\pi\left(\dfrac15-\dfrac83+16\right)\\ &=\boxed{\boxed{\dfrac{203}{15}\pi}} \end{aligned}

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

No.

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{aligned} V&=\pi\displaystyle\intop_0^{\pi}(x\sin x)^2\ dx\\ &=\pi\displaystyle\intop_0^{\pi}x^2\sin^2 x\ dx\\ &=\pi\displaystyle\intop_0^{\pi}x^2\left(\dfrac12-\dfrac12\cos2x\right)\ dx\\ &=\pi\displaystyle\intop_0^{\pi}\dfrac12x^2-\dfrac12x^2\cos2x\ dx\\ \end{aligned}

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

\begin{aligned} V&=\pi\left[\dfrac12\cdot\dfrac13x^3+\left(-\dfrac12x^2\right)\left(\dfrac12\sin2x\right)-(-x)\left(-\dfrac14\cos2x\right)+(-1)\left(-\dfrac18\sin2x\right)\right]_0^{\pi}\\ &=\pi\left[\dfrac16x^3-\dfrac14x^2\sin2x-\dfrac14x\cos2x+\dfrac18\sin2x\right]_0^{\pi}\\ &=\pi\left[\left(\dfrac16\pi^3-\dfrac14\pi^2\sin2\pi-\dfrac14\pi\cos2\pi+\dfrac18\sin2\pi\right)-\left(\dfrac16(0)^3-\dfrac14(0)^2\sin2(0)-\dfrac14(0)\cos2(0)+\dfrac18\sin2(0)\right)\right]\\ &=\pi\left[\left(\dfrac16\pi^3-\dfrac14\pi^2(0)-\dfrac14\pi(1)+\dfrac18(0)\right)-0\right]\\ &=\pi\left(\dfrac16\pi^3-\dfrac14\pi\right)\\ &=\dfrac{\pi^2}6\left(\pi^2-\dfrac64\right)\\ &=\dfrac{\pi^2}6\left(\pi^2-\dfrac32\right)\\ \end{aligned}

Jadi, a=\dfrac32

No.

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

brainly.co.id

\begin{aligned}L&=\displaystyle\intop_1^32x\ dx\\&=\left.x^2\right|_1^3\\&=3^2-1^2\\&=9-1\\&=\boxed{\boxed{8}}\end{aligned}

\begin{aligned}V&=\pi\displaystyle\intop_1^3(2x)^2\ dx\\&=\pi\displaystyle\intop_1^34x^2\ dx\\&=\pi\left[\dfrac43x^3\right]_1^3\\&=\pi\left[\dfrac43(3)^3-\dfrac43(1)^3\right]\\&=\pi\left[\dfrac43(27)-\dfrac43(1)\right]\\&=\pi\left[36-\dfrac43\right]\\&=\boxed{\boxed{34\dfrac23\pi}}\end{aligned}

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