高専の数学解説

第5章 5.43

I=∫02π(xcos⁡θ+ysin⁡θ−θ)2 dθI = \int_0^{2\pi} (x\cos\theta + y\sin\theta - \theta)^2\, d\theta

=∫02π(x2cos⁡2θ+xycos⁡θsin⁡θ−xθcos⁡θ+xycos⁡θsin⁡θ= \int_0^{2\pi} (x^2\cos^2\theta + xy\cos\theta\sin\theta - x\theta\cos\theta + xy\cos\theta\sin\theta

+y2sin⁡2θ−yθsin⁡θ−xθcos⁡θ−yθsin⁡θ+θ2) dθ+ y^2\sin^2\theta - y\theta\sin\theta - x\theta\cos\theta - y\theta\sin\theta + \theta^2)\,d\theta

=∫02π(x2cos⁡2θ+y2sin⁡2θ+θ2+2xycos⁡θsin⁡θ−2xθcos⁡θ−2yθsin⁡θ) dθ= \int_0^{2\pi} (x^2\cos^2\theta + y^2\sin^2\theta + \theta^2 + 2xy\cos\theta\sin\theta - 2x\theta\cos\theta - 2y\theta\sin\theta)\,d\theta

=[x2(12θ−14sin⁡2θ)+y2(12θ+14sin⁡2θ)+13θ3]02π−2x[θsin⁡θ]02π+∫02πsin⁡θ dθ= \left[x^2\left(\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right) + y^2\left(\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right) + \frac{1}{3}\theta^3\right]_0^{2\pi} - 2x[\theta\sin\theta]_0^{2\pi} + \int_0^{2\pi} \sin\theta\, d\theta

−2y[θ(−cos⁡θ)]02π−∫02πcos⁡θ dθ- 2y[\theta(-\cos\theta)]_0^{2\pi} - \int_0^{2\pi} \cos\theta\, d\theta

=πx2+πy2+83π3+4πy= \pi x^2 + \pi y^2 + \frac{8}{3}\pi^3 + 4\pi y

=πx2+83π3+{π(y+2)2−4π}= \pi x^2 + \frac{8}{3}\pi^3 + \{\pi(y+2)^2 - 4\pi\}

これから

x=0x = 0、y=−2y = -2 のとき最小となり

I=83π3−4πとなるI = \frac{8}{3}\pi^3 - 4\pi \quad \text{となる}

解説: r31bn1z