高専の数学解説

第8章 8.21

(1)

x2+3y2x^2+3y^2 D:x2+y2≤1D: x^2+y^2 \leq 1

x=rcos⁡θx=r\cos\theta y=rsin⁡θy=r\sin\theta とおくと 0≤r≤10 \leq r \leq 1 0≤θ≤2π0 \leq \theta \leq 2\pi

∫02π∫01r3(cos⁡2θ+3sin⁡2θ)drdθ=∫02π14(cos⁡2θ+3sin⁡2θ)dθ=∫02π14(1+2sin⁡2θ)dθ\int_0^{2\pi}\int_0^1 r^3(\cos^2\theta+3\sin^2\theta)drd\theta=\int_0^{2\pi}\frac{1}{4}(\cos^2\theta+3\sin^2\theta)d\theta=\int_0^{2\pi}\frac{1}{4}(1+2\sin^2\theta)d\theta

=∫02π14(1+2(1−cos⁡2θ)2)dθ=14[θ+θ−12sin⁡2θ]02π=π=\int_0^{2\pi}\frac{1}{4}\left(1+\frac{2(1-\cos2\theta)}{2}\right)d\theta=\frac{1}{4}\left[\theta+\theta-\frac{1}{2}\sin2\theta\right]_0^{2\pi}=\pi

(2)

x2y1+x2+y2\dfrac{x^2y}{1+x^2+y^2} D:x2+y2≤1D: x^2+y^2 \leq 1 x≥0x \geq 0 y≥0y \geq 0

x=rcos⁡θx=r\cos\theta y=rsin⁡θy=r\sin\theta とおくと 0≤r≤10 \leq r \leq 1 0≤θ≤π20 \leq \theta \leq \dfrac{\pi}{2}

∫01∫0π2r4cos⁡2θ⋅sin⁡θ1+r2dθdr\int_0^1\int_0^{\frac{\pi}{2}}\frac{r^4\cos^2\theta\cdot\sin\theta}{1+r^2}d\theta dr

cos⁡θ=t\cos\theta=tとおくと sin⁡θdθ=dt\sin\theta d\theta=dt θ:0→π2\theta:0 \to \dfrac{\pi}{2} t:1→0t:1 \to 0 これから

=∫01∫01r41+r2t2dtdr=∫0113⋅r41+r2dr=∫0113(r2−1+11+r2)dr=\int_0^1\int_0^1\frac{r^4}{1+r^2}t^2 dtdr=\int_0^1\frac{1}{3}\cdot\frac{r^4}{1+r^2}dr=\int_0^1\frac{1}{3}\left(r^2-1+\frac{1}{1+r^2}\right)dr

=[13(13r3−r+tan⁡−1r)]01=13(π4−23)=\left[\frac{1}{3}\left(\frac{1}{3}r^3-r+\tan^{-1}r\right)\right]_0^1=\frac{1}{3}\left(\frac{\pi}{4}-\frac{2}{3}\right)

(3)

cos⁡(x2+y2)\cos(x^2+y^2) D:x2+y2≤π2D: x^2+y^2 \leq \dfrac{\pi}{2}

x=rcos⁡θx=r\cos\theta y=rsin⁡θy=r\sin\theta とおくと 0≤r≤π20 \leq r \leq \sqrt{\dfrac{\pi}{2}} 0≤θ≤2π0 \leq \theta \leq 2\pi

∫02π∫0π2r⋅cos⁡r2 drdθ=∫02π[12sin⁡r2]0π2dθ=∫02π12dθ=π\int_0^{2\pi}\int_0^{\sqrt{\frac{\pi}{2}}} r\cdot\cos r^2\,drd\theta=\int_0^{2\pi}\left[\frac{1}{2}\sin r^2\right]_0^{\sqrt{\frac{\pi}{2}}}d\theta=\int_0^{2\pi}\frac{1}{2}d\theta=\pi

(4)

9−x2−y2D:x2+y2≤3x\sqrt{9-x^2-y^2} \qquad D : x^2+y^2 \le 3x

x=rcos⁡θx = r\cos\theta y=rsin⁡θy = r\sin\theta とおくと 0≤r≤3cos⁡θ0 \le r \le 3\cos\theta 0≤θ≤π0 \le \theta \le \pi

∫0π∫03cos⁡θr9−r2 dr dθ=∫0π[−13(9−r2)3]03cos⁡θdθ=∫0π−9sin⁡3θ+9 dθ=18∫0π21−sin⁡3θ dθ\int_0^{\pi}\int_0^{3\cos\theta} r\sqrt{9-r^2}\,dr\,d\theta = \int_0^{\pi}\left[-\frac{1}{3}\sqrt{(9-r^2)^3}\right]_0^{3\cos\theta} d\theta = \int_0^{\pi} -9\sin^3\theta+9\,d\theta = 18\int_0^{\frac{\pi}{2}} 1-\sin^3\theta\,d\theta

=18(π2−23)=9π−12= 18\left(\frac{\pi}{2}-\frac{2}{3}\right) = 9\pi - 12

(5)

11+x2+y2D:x2+y2≤1\frac{1}{\sqrt{1+x^2+y^2}} \qquad D : x^2+y^2 \le 1

x=rcos⁡θx = r\cos\theta y=rsin⁡θy = r\sin\theta とおくと 0≤r≤10 \le r \le 1 0≤θ≤2π0 \le \theta \le 2\pi

∫02π∫01r1+r2 dr dθ=∫02π[1+r2]01dθ=∫02π2−1 dθ=[(2−1)θ]02π=2π(2−1)\int_0^{2\pi}\int_0^{1} \frac{r}{\sqrt{1+r^2}}\,dr\,d\theta = \int_0^{2\pi}\left[\sqrt{1+r^2}\right]_0^{1} d\theta = \int_0^{2\pi} \sqrt{2} - 1\,d\theta = \left[(\sqrt{2}-1)\theta\right]_0^{2\pi} = 2\pi(\sqrt{2}-1)

解説: r31bn1z