9.5-bo'lim : Parametrik tenglamalar bilan sirt maydoni
5. Parametrik egri chiziqni berilgan o'q atrofida aylantirish natijasida olingan ob'ektning sirt maydonini beradigan integralni o'rnating, lekin baholamang. Egri chiziq \(t\) ning berilgan diapazoni uchun aynan bir marta chiqadi deb taxmin qilishingiz mumkin.
\(\displaystyle x = 1 + 3{t^2}\hspace{0.25in}y = \sin \left( {2t} \right)\cos \left( {\frac{1}{4}t} \right)\hspace{0.25in}0 \le t \le \frac{1}{2}\)-ni \(y\) o'qi atrofida aylantiring.
Yechim
Bu erda bizga kerak bo'lgan birinchi narsa quyidagi ikkita hosiladir.
\[\frac{{dx}}{{dt}} = 6t\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 2\cos \left( {2t} \right)\cos \left( {\frac{1}{4}t} \right) - \frac{1}{4}\sin \left( {2t} \right)\sin \left( {\frac{1}{4}t} \right)\]2-qadam
Ushbu muammo uchun bizga \(ds\) kerak bo'ladi.
\[ds = \sqrt {{{\left[ {6t} \right]}^2} + {{\left[ {2\cos \left( {2t} \right)\cos \left( {\frac{1}{4}t} \right) - \frac{1}{4}\sin \left( {2t} \right)\sin \left( {\frac{1}{4}t} \right)} \right]}^2}} \,dt\]3-qadam
U holda sirt maydoni uchun integral,
\[SA = \int_{{}}^{{}}{{2\pi x\,ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{0}^{{\frac{1}{2}}}{{2\pi \left( {1 + 3{t^2}} \right)\sqrt {36{t^2} + {{\left( {2\cos \left( {2t} \right)\cos \left( {\frac{1}{4}t} \right) - \frac{1}{4}\sin \left( {2t} \right)\sin \left( {\frac{1}{4}t} \right)} \right)}^2}} \,dt}}}}\]Sirt maydoni uchun formula bilan ehtiyot bo'lishni unutmang! Amaldagi formula biz aylanayotgan o'qga bog'liq.