Question

Необходимо вычислить длину дуги и объем тела, полученного вращение вокруг оси Оx

Answer 1

Ответ:

${}\ \ y=x^2-1\ \ ,\ \ OX:y=0\ \ ,\\\\x^2-1=0\ \ ,\ \ x^2=1\ \ ,\ \ x=\pm 1\\\\l=\int\limits^1_{-1} \, \sqrt{1+(y')^2}\, dx=\int\limits^1_{-1} \, \sqrt{1+(2x)^2}\, dx=\int\limits^1_{-1} \, \sqrt{1+4x^2}\, dx=(\star )\\\\\\Q=\int \sqrt{1+4x^2}\, dx=\Big[\ u=\sqrt{1+4x^2}\ ,\ du=\dfrac{8x\, dx}{2\sqrt{1+4x^2}}=\dfrac{4x\, dx}{\sqrt{1+4x^2}}\ ,\\\\\\dv=dx\ \ ,\ \ v=x\ \Big]=uv-\int v\, du=x\cdot \sqrt{1+4x^2}-\int \dfrac{4x^2\, dx}{\sqrt{1+4x^2}}=$

$=x\cdot \sqrt{1+4x^2}-\int \dfrac{(1+4x^2)-1}{\sqrt{1+4x^2}}\, dx=x\cdot \sqrt{1+4x^2}-\int \Big(\dfrac{1+4x^2}{\sqrt{1+4x^2}}-\dfrac{1}{\sqrt{1+4x^2}}\Big)\, dx=\\\\\\=x\cdot \sqrt{1+4x^2}-\int \sqrt{1+4x^2}\, dx+\int \dfrac{dx}{\sqrt{1+4x^2}}=\\\\\\=x\cdot \sqrt{1+4x^2}-\int \sqrt{1+4x^2}\, dx+\dfrac{1}{2}\int \dfrac{d(2x)}{\sqrt{1+(2x)^2}}=\\\\\\=x\cdot \sqrt{1+4x^2}-Q+\dfrac{1}{2}\cdot ln\Big|\, 2x+\sqrt{1+4x^2}\, \Big|+C\ \ \ \Rightarrow$

$2Q=x\cdot \sqrt{1+4x^2}+\dfrac{1}{2}\cdot ln\Big|\, 2x+\sqrt{1+4x^2}\, \Big|+C\\\\\\Q=\dfrac{1}{2}\, x\cdot \sqrt{1+4x^2}+\dfrac{1}{4}\cdot ln\Big|\, 2x+\sqrt{1+4x^2}\, \Big|+\dfrac{C}{2}$

$(\star )=\int\limits^1_{-1}\, \sqrt{1+4x^2}\, dx=\Big(\dfrac{1}{2}\, x\cdot \sqrt{1+4x^2}+\dfrac{1}{4}\cdot ln\Big|\, 2x+\sqrt{1+4x^2}\, \Big|\Big)\Big|_{-1}^1=\\\\\\=\dfrac{1}{2}\cdot \sqrt5+\dfrac{1}{4}\cdot ln\, \Big|\, 2+\sqrt5\, \Big|+\dfrac{1}{2}\cdot \sqrt5-\dfrac{1}{4}\cdot ln\, \Big|\, -2+\sqrt5\, \Big|=\\\\\\=\sqrt5+\dfrac{1}{4}\cdot ln\Big(\dfrac{2+\sqrt5}{\sqrt5-2}\Big)\ ;$

$l=\sqrt5+\dfrac{1}{4}\cdot ln\Big(\dfrac{2+\sqrt5}{\sqrt5-2}\Big)$