Question

Помогите пожалуйста решить с подобным решением ​

Answer 1

Объяснение:

$x^2-2x+y^2=0\ \ \ \ \ x^2-6x+y^2=0\ \ \ \ y=x\ \ \ \ y=0.$

Перейдемо до полярних координат:

$x=r*cos\phi\ \ \ \ y=r*sin\phi\ \ \ \ \Rightarrow\\(r*cos\phi)^2-2*r*cos\phi+(r*sin\phi)^2=0\\r^2*cos^2\phi-2r*cos\phi+r^2*sin^2\phi=0\\r^2*(cos^2\phi+sin^2\phi)=2r*cos\phi\\r^2=2r*cos\phi\ |:r\\\boxed {r=2cos\phi}.$

$(r*cos\phi)^2-6*r*sin\phi+(r*sin\phi)^2=0\\r^2*cos^2\phi-6r*sin\phi+r^2*sin^2\phi=0\\r^2*(cos^2\phi+sin^2\phi)=6r*sin\phi\\r^2=2r*sin\phi\ |:r\\\boxed {r=6sin\phi}.$

$rsin\phi=rcos\phi\ |:rcos\phi\\tg\phi=1\\\boxed {\phi=\frac{\pi }{4}}.$

$rsin\phi=0|:r\\sin\phi=0\\\boxed {\phi=0}.$

$\int\limits^\frac{\pi }{4} _ {0} \, d\phi\int\limits^{6sin\phi}_{2cos\phi} {r} \, dr=\int\limits^\frac{\pi }{4} _0 \, d\phi\frac{r^2}{2}\ |_{2cos\phi}^{6sin\phi}=\frac{1}{2}\int\limits^\frac{\pi }{4} _ {0} \,((6sin\phi)^2-(2cos\phi)^2) d\phi=\\ =\frac{1}{2}*\int\limits^\frac{\pi }{4} _0 {(36sin^2\phi-4cos^2\phi)} \, d\phi =\frac{4}{2}* \int\limits^\frac{\pi }{4} _0 {(9sin^2\phi-cos^2\phi)} \, d\phi=$

$=2*\int\limits^\frac{\pi }{4} _0 {9sin^2\phi} \, d\phi -2*\int\limits^\frac{\pi }{4} _0 {cos^2\phi} \, d\phi .$

$\int\limits sin^2\phi \, d\phi=\int\frac{1-cos2\phi}{2} d\phi=\frac{1}{2}*\int(1-cos2\phi)d\phi=\frac{1}{2}\int d\phi -\frac{1}{4} \int cos2\phi d2\phi=\\=\frac{\phi}{2}-\frac{sin2\phi}{4} .\\2*9*(\frac{\phi}{2}-\frac{sin2\phi}{4}) \ |_0^\frac{\pi }{4} =18*(\frac{\pi }{8} -\frac{1}{4}) =\frac{9\pi }{4}-\frac{9}{2} .$

$\int cos^2\phid\phi=\int \frac{1+cos2\phi}{2} =\frac{1}{2} \int (1+cos2\phi)d\phi=\frac{1}{2} \int d\phi+\frac{1}{4} \int cos2\phi d\phi =\\=\frac{\phi}{2} +\frac{sin2\phi}{4} .\\2*(\frac{\phi}{2} +\frac{sin2\phi}{4})\ |_0^\frac{\pi }{4} =2*(\frac{\pi }{8} +\frac{1}{4})=\frac{\pi }{4} +\frac{1}{2}.$

$\frac{9\pi }{4} -\frac{9}{2}-\frac{\pi }{4} -\frac{1}{2} =\frac{8}{4}\pi -\frac{3}{4}=2\pi -0,75\approx 5,53.$