Question

Помогите пожалуйста ​

Answer 1

Ответ:

$6)\ \ r=2\, (1+sin\varphi )\ \ ,\ \ \ -\dfrac{\pi}{6}\leq \varphi \leq 0\\\\l=\int\limits^{\varphi _2}_{\varphi _1}\, \sqrt{r^2(\varphi )+(r'(\varphi ))^2}\, d\varphi$

$r'(\varphi )=2\cdot cos\varphi \ \ ,\ \ r^2(\varphi )+(r'(\varphi ))^2=4(1+sin\varphi )^2+4\, cos^2\varphi =\\\\=4(1+2sin\varphi +sin^2\varphi )+4\, cos^2\varphi =4+8sin\varphi +\underbrace {4sin^2\varphi +4cos^2\varphi }_{4}=\\\\=8+8sin\varphi =8\, (1+sin\varphi )=8\, (1+cos(\frac{\pi}{2}-\varphi ))=8\cdot 2cos^2(\dfrac{\pi}{4}-\dfrac{\varphi }{2})\ ;$

$l=\int\limits^0_{-\pi /6}\, \sqrt{16\, cos^2(\dfrac{\pi}{4}-\dfrac{\varphi }{2})}\, d\varphi =4\int\limits^0_{-\pi /6}\, cos(\dfrac{\pi}{4}-\dfrac{\varphi }{2})\, d\varphi =-4\cdot 2sin(\dfrac{\pi}{4}-\dfrac{\varphi }{2})\Big|^0_{-\frac{\pi}{6}}=\\\\\\=-8\cdot (sin\dfrac{\pi}{4}-sin(\dfrac{\pi}{4}+\dfrac{\pi}{12}))=-8\cdot (\dfrac{\sqrt2}{2}-sin\dfrac{\pi}{3})=-8\cdot (\dfrac{\sqrt2}{2}-\dfrac{\sqrt3}{2})=\\\\\\=-\dfrac{8}{2}\cdot (\sqrt2-\sqrt3)=4\cdot (\sqrt3-\sqrt2)\approx 1,2713$

$7)\ \ y=e^{x}\ \ ,\ \ y=e^{2x}\ \ ,\ \ x=1\\\\V_{ox}=\pi \int\limits^a_b\, f^2(x)\, dx\\\\\\V_{ox}=\int\limits^1_0\, (e^{2x}-e^{x})\, dx=\Big(\dfrac{1}{2}\cdot e^{2x}-e^{x}\Big)\Big|_0^1=\Big(\dfrac{1}{2}e^2-e\Big)-\Big(\dfrac{1}{2}\cdot 1-1\Big)=\\\\\\=\dfrac{1}{2}\cdot e^2-e+\dfrac{1}{2}=\dfrac{1}{2}\cdot \Big(e^2+1\Big)-e$