Double Integrals & Order of Integration

A rectangle: no curves to intersect.

Either order works; use $dy\,dx$.

$\displaystyle\int_0^2\int_1^3 6xy^2\,dy\,dx$.

$\displaystyle\int_1^3 6xy^2\,dy = 2xy^3\Big|_1^3 = 52x$.

$\displaystyle\int_0^2 52x\,dx = 26x^2\Big|_0^2 = 104$.

$x^2 = 2x$ at $x = 0$ and $x = 2$. For $0 < x < 2$, the line $y = 2x$ is on top.

$dy\,dx$: top $y = 2x$, bottom $y = x^2$.

$\displaystyle\int_0^2\int_{x^2}^{2x} xy\,dy\,dx$.

$\dfrac{xy^2}{2}\Big|_{x^2}^{2x} = \dfrac{x(4x^2 - x^4)}{2} = 2x^3 - \dfrac{x^5}{2}$.

$\displaystyle\int_0^2 \left(2x^3 - \dfrac{x^5}{2}\right)\,dx = 8 - \dfrac{16}{3} = \dfrac{8}{3}$.

$y^2 - 1 = 1 - y^2$ at $y = \pm 1$. Left curve $x = y^2 - 1$, right curve $x = 1 - y^2$.

$dx\,dy$. (In $dy\,dx$ the top and bottom curves change at $x = 0$, so it would split.)

$\displaystyle\int_{-1}^{1}\int_{y^2 - 1}^{1 - y^2} y^2\,dx\,dy$.

$y^2x\Big|_{y^2 - 1}^{1 - y^2} = 2y^2 - 2y^4$.

$\displaystyle\int_{-1}^{1} (2y^2 - 2y^4)\,dy = \dfrac{4}{3} - \dfrac{4}{5} = \dfrac{8}{15}$.

$0 \le y \le 3$, $\tfrac{y}{3} \le x \le 1$: the triangle with vertices $(0, 0)$, $(1, 0)$, $(1, 3)$.

$0 \le x \le 1$.

$x = \tfrac{y}{3}$ is $y = 3x$, so $0 \le y \le 3x$: $\displaystyle\int_0^1\int_0^{3x} e^{x^2}\,dy\,dx$.

$\displaystyle\int_0^{3x} e^{x^2}\,dy = 3xe^{x^2}$.

$\displaystyle\int_0^1 3xe^{x^2}\,dx = \tfrac32 e^{x^2}\Big|_0^1 = \tfrac32(e - 1)$.

$0 \le y \le 8$, $\sqrt[3]{y} \le x \le 2$: bounded by $y = x^3$, $x = 2$ and $y = 0$.

$0 \le x \le 2$.

$x = \sqrt[3]{y}$ is $y = x^3$, so $0 \le y \le x^3$: $\displaystyle\int_0^2\int_0^{x^3} \dfrac{1}{x^4 + 1}\,dy\,dx$.

$\displaystyle\int_0^{x^3} \dfrac{1}{x^4 + 1}\,dy = \dfrac{x^3}{x^4 + 1}$.

$u = x^4 + 1$: $\tfrac14\log(x^4 + 1)\Big|_0^2 = \tfrac14\log 17$.

$0 \le x \le \sqrt\pi$, $x \le y \le \sqrt\pi$: the triangle with vertices $(0, 0)$, $(0, \sqrt\pi)$, $(\sqrt\pi, \sqrt\pi)$.

$0 \le y \le \sqrt\pi$.

$0 \le x \le y$: $\displaystyle\int_0^{\sqrt\pi}\int_0^{y} \sin(y^2)\,dx\,dy$.

$\displaystyle\int_0^{y} \sin(y^2)\,dx = y\sin(y^2)$.

$u = y^2$: $-\tfrac12\cos(y^2)\Big|_0^{\sqrt\pi} = -\tfrac12(\cos\pi - 1) = 1$.

Piece 1: $0 \le y \le 1$, $0 \le x \le 2y$. Piece 2: $1 \le y \le 3$, $0 \le x \le 3 - y$.

They meet along $y = 1$ for $0 \le x \le 2$. Together: the triangle with vertices $(0, 0)$, $(2, 1)$, $(0, 3)$.

$0 \le x \le 2$.

Bottom $x = 2y$, so $y = \tfrac{x}{2}$; top $x = 3 - y$, so $y = 3 - x$. Area check with $f = 1$: $1 + 2 = 3$ both ways.

$0 \le y \le 1$, $y^2 \le x \le 2 - y$: left curve $x = y^2$, right line $x = 2 - y$, meeting at $(1, 1)$.

One piece, bounded by $y = 0$, the curve $y = \sqrt{x}$ from $(0, 0)$ to $(1, 1)$, and the line $y = 2 - x$ from $(1, 1)$ to $(2, 0)$.

$0 \le x \le 2$.

The top changes at $x = 1$: $y = \sqrt{x}$ for $0 \le x \le 1$, then $y = 2 - x$ for $1 \le x \le 2$. Area check: $\tfrac76$ both ways.

The plane meets $z = 0$ along $x + 2y = 4$: base triangle $x, y \ge 0$, $x + 2y \le 4$.

Height $4 - x - 2y$ (bottom $z = 0$).

$V = \displaystyle\int_0^4\int_0^{(4 - x)/2} (4 - x - 2y)\,dy\,dx$.

Inner: $(4 - x)y - y^2\Big|_0^{(4 - x)/2} = \dfrac{(4 - x)^2}{4}$. Outer: $\displaystyle\int_0^4 \dfrac{(4 - x)^2}{4}\,dx = \dfrac{16}{3}$.

Between the parabola $y = x^2$ and the line $y = 1$, for $-1 \le x \le 1$.

Height $x^2 + y^2$ (bottom $z = 0$).

$V = \displaystyle\int_{-1}^{1}\int_{x^2}^{1} (x^2 + y^2)\,dy\,dx$.

Inner: $x^2(1 - x^2) + \dfrac{1 - x^6}{3}$. Outer: $2\left(\dfrac{2}{15} + \dfrac{2}{7}\right) = \dfrac{88}{105}$.

Top above bottom when $6 - x - y \ge x + y$, that is $x + y \le 3$: base triangle $x, y \ge 0$, $x + y \le 3$.

Height $(6 - x - y) - (x + y) = 6 - 2x - 2y$.

$V = \displaystyle\int_0^3\int_0^{3 - x} (6 - 2x - 2y)\,dy\,dx$.

Inner: $(6 - 2x)(3 - x) - (3 - x)^2 = (3 - x)^2$. Outer: $\displaystyle\int_0^3 (3 - x)^2\,dx = 9$.

$-1 \le x \le 2$, $0 \le y \le 4$: a $3 \times 4$ rectangle.

Integrand $1$: the area of the rectangle.

$3 \times 4 = 12$.

$-3 \le x \le 3$, $0 \le y \le \sqrt{9 - x^2}$: the upper half of the disk of radius $3$.

Integrand $1$: its area.

$\tfrac12\pi(3)^2 = \tfrac{9\pi}{2}$.

The disk of radius $2$.

$z = \sqrt{4 - x^2 - y^2}$ is the upper hemisphere of radius $2$.

$\tfrac12 \cdot \tfrac43\pi(2)^3 = \tfrac{16\pi}{3}$.