Krivuljni integral prve vrste na Geodetskom fakultetu
Na pismenom ispitu iz Vektorske analize 20. 6. 2011. je bio sljedeći zadatak:
Izračunajte
gdje je k prvi zavoj cilindrične zavojnice
Rješenje.
Izračunajmo prvo koliki je ds. Kako je
to je
![\begin{array}{rcl} ds &=& \displaystyle \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\, dt=\sqrt{(-a\sin t)^2+(a\cos t)^2+a^2}\, dt\\[15pt] &=& \displaystyle \sqrt{a^2\sin^2 t + a^2\cos^2 t + a^2}\, dt = a\sqrt{2}\, dt \end{array}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_us33AXTjNwXhH6CQuX9egTyAWEjzAcKCU1xpR0AEnPN99t_evNmVKOI9RzfBFWKhyiYgayoFUegpn2c2xX1sWV1pyTH2C3iq9PGzMolWoUajGtYsg5cogjw5QGp6j6aHT1IO8xK5xxdDvGXTufNEtPQhbWIu05MWsppydDcwpGOJ2wpEdOF1GcDPlw-25RfekHxoF-_ZF9G4320x75KPErFNj2l0X_f3uyuPqHJfu58rVSgi71LS7NmC12dqFU4fibtihPwAy0M-h9djiE6RbWvNrsdJ1cCH0OFXZdydXfJCjULGKc5HP8DIXdbB65OiLfcFA47ndTAEIsIe8NbbBQ_0MAngvy6Oz33NkrMf4ER8_oyMHLiJtCcOJqnXK63u8443O5vD7BX3oTGt5puMohyW49TuapY_kgowdT5f3e6fmfkNPI6PVVTzsNfymeBkkJdYBMrlJtGJUoWlMeEA2Khu-LmD3r8rIzVexziOfnUrhv-YTope58cMNT2zy-ITrBMTtZdV3WXI83nf3Q6o0OrZV5Gs5fkkpzqPt8TXb6L2epeF9_xuQR_a2ujbqrG8ukByEfCDUhoaMmETlXHXDF_h4rW7k=s0-d "\begin{array}{rcl} ds &=& \displaystyle \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\, dt=\sqrt{(-a\sin t)^2+(a\cos t)^2+a^2}\, dt\\[15pt] &=& \displaystyle \sqrt{a^2\sin^2 t + a^2\cos^2 t + a^2}\, dt = a\sqrt{2}\, dt \end{array}")
pa je, jer su granice
![\begin{array}{rcl} \displaystyle \int_k \frac{z^2}{x^2+y^2}\, ds &=& \displaystyle \int_0^{2\pi}\frac{(at)^2}{(a\cos t)^2 + (a\sin t)^2}\, a\sqrt{2}\, dt\\[20pt] &=& \displaystyle a\sqrt{2}\int_0^{2\pi} t^2\, dt = \frac{8\sqrt{2} \pi^3 a}{3} \end{array}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uJpsMdWPQUu8XBLZocHcv00SOqtMy7z5l1FyPeitJuvV2aQsv_HVu9jkiw4RYu_9TScH6vgwgqbJo97Y8ls4A45bewpyF0tWHhEi4P6tHGW1_5QkkJcCCr1PiylsQ3hgDSCFo5QH2DiPrIf9Bl5rh7mSsF5ld4nVr5y2Vbarr_uqzwGYp_LBgLzDgELJ-xKyOkUF4X8rap3LLN4DmMXbJjRRFD7p2Ffua6WJIWo0G2K9LhpWh44sfcuJ6tVMiiO6_3qBfD8hPXC-OocjurP-IfC4yERreYmWpkrNqlDVu7v-GRn6OGetVp42cADeKcqsrteBPrpRfHnbfUgQbU5h7b9yNU5xztewdvSHvvsG1FJf3gD5DCgQHhJpiLPzBFzUzvKJx-VGYxpTbkg0nHN7EFIfEhVhalBnPw-IXm7UCDQ9AnCAhJflWbTJMs-T27ySoYUB9h_XfqMzrJJfpFk72PzduqmIB6Ffsh2S6v_cLkYu2i3A4q7N412FK9os3dJW0V6FQ_QrX0sH8elP_6or62BFCC8vGfzh9YvJx03sbyoZijv0nxWbhtduQPI2x2ISzxjTkgRmcetT7YUVUfS9y73wZFYwmKMhqIxuZJDVhYZIwCKgVS2RrA3xB0xa1RaBQDStdwAtYCVaS73i018KWWjw=s0-d "\begin{array}{rcl} \displaystyle \int_k \frac{z^2}{x^2+y^2}\, ds &=& \displaystyle \int_0^{2\pi}\frac{(at)^2}{(a\cos t)^2 + (a\sin t)^2}\, a\sqrt{2}\, dt\\[20pt] &=& \displaystyle a\sqrt{2}\int_0^{2\pi} t^2\, dt = \frac{8\sqrt{2} \pi^3 a}{3} \end{array}")
