【微分積分】8-1-2 一般集合上の2重積分｜問題集

\(\displaystyle =\int_{-1}^1xdx\int_0^31dy\)
\(\displaystyle =\left[\frac{x^2}{2}\right]_{-1}^{1}・[y]_0^3\)
\(=0・3\)
\(=0\)

\(\displaystyle =\int_0^1\int_0^x(2x+3y)dydx\)
\(\displaystyle =\int_0^1\left[2xy+\frac{3y^2}{2}\right]_0^xdx\)
\(\displaystyle =\int_0^1\frac{7x^2}{2}dx\)
\(\displaystyle =\left[\frac{7}{6}x^3\right]_0^1\)
\(\displaystyle =\frac{7}{6}\)

\(\displaystyle =\int_0^1\int_{y^2}^y(1+x+xy)dxdy\)
\(\displaystyle =\int_0^1\left[x+\frac{x^2}{2}+\frac{x^2y}{2}\right]_{y^2}^ydy\)
\(\displaystyle =\int_0^1\left(y-\frac{y^2}{2}+\frac{y^3}{2}-\frac{y^4}{2}-\frac{y^5}{2}\right)dy\)
\(\displaystyle =\left[\frac{y^2}{2}-\frac{y^3}{6}+\frac{y^4}{8}-\frac{y^5}{10}-\frac{y^6}{12}\right]_0^1\)
\(\displaystyle =\frac{11}{40}\)

\(\displaystyle =\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\sin(x+y)dydx\)
\(\displaystyle =\int_0^\frac{\pi}{2}[-\cos(x+y)]_0^\frac{\pi}{2}dx\)
\(\displaystyle =\int_0^\frac{\pi}{2}(\sin x+\cos x)dx\)
\(\displaystyle =[-\cos x+\sin x]_0^\frac{\pi}{2}\)
\(=2\)

\(\displaystyle =\int_0^1\left(\int_0^xx^3ydy\right)dx\)
\(\displaystyle =\int_0^1\left[\frac{x^3y^2}{2}\right]_0^xdx\)
\(\displaystyle =\int_0^1\frac{x^5}{2}dx\)
\(\displaystyle =\left[\frac{x^6}{12}\right]_{0}^{1}\)
\(\displaystyle =\frac{1}{12}\)

\(\displaystyle =\int_{-1}^1\left(\int_0^3x^2dy\right)dx\)
\(\displaystyle =\int_{-1}^1[x^2y]_0^3dx\)
\(\displaystyle =\int_{-1}^13x^2dx\)
\(\displaystyle =[x^3]_{-1}^1\)
\(=2\)

\(\displaystyle =\int_0^1\left(\int_0^xe^{x+y}dy\right)dx\)
\(\displaystyle =\int_0^1[e^{x+y}]_0^xdx\)
\(\displaystyle =\int_0^1(e^{2x}-e^x)dx\)
\(\displaystyle =\left[\frac{e^{2x}}{2}-e^x\right]_0^1\)
\(\displaystyle =\frac{e^2}{2}-e+\frac{1}{2}\)

\(\displaystyle =\int_0^1\left(\int_{y^2}^y\sqrt{xy}dx\right)dy\)
\(\displaystyle =\int_0^1\left[\sqrt{y}・\frac{2}{3}x^{\frac{3}{2}}\right]_{y^2}^ydy\)
\(\displaystyle =\frac{2}{3}\int_0^1(y^2-y^\frac{7}{2})dy\)
\(\displaystyle =\frac{2}{3}\left[\frac{y^3}{3}-\frac{2}{9}y^\frac{9}{2}\right]_0^1\)
\(\displaystyle =\frac{2}{27}\)

\(y^2=2x\)と\(y^2=8-2x\)の交点を求めると、\(x=2,y=\pm2\)となる。
\(\displaystyle D=\left\{(x,y)|-2\leqq y\leqq 2,\frac{y^2}{2}\leqq x\leqq \frac{8-y^2}{2}\right\}\)
よって、
\(\displaystyle \iint_D(4-y^2)dxdy\)
\(\displaystyle =\int_{-2}^2\left(\int_\frac{y^2}{2}^\frac{8-y^2}{2}(4-y^2)dx\right)dy\)
\(\displaystyle =\int_{-2}^2\left[4x-y^2x\right]_\frac{y^2}{2}^\frac{8-y^2}{2}dy\)
\(\displaystyle =\int_{-2}^2(y^4-8y^2+16)dy\)
\(\displaystyle =2\left[\frac{y^5}{5}-\frac{8}{3}y^3+16y\right]_0^2\)
\(\displaystyle =\frac{512}{15}\)