3.1 Double Integrals

(a) $20$ (b) $\pi$ (c) $9\pi$

Be careful to match each integration variable with its own limits. Remember that the integral with respect to $x$ treats $y$ as a constant and the integral with respect to $y$ treats $x$ as a constant.

(a) $108 y^3$ (b) $48x^2$ (c), (d) $432$ (e) $648$

(a) $\frac{1}{3}\big(a^3b+ab^3\big)$ (b) $\frac{a^2b}{6}-\frac{ab^2}{2}$ (c) $\frac{3}{56}$ (d) $\frac{1}{2}(1-\cos 1)$ (e) $\frac{1}{2}(e-2)$

(f) $\frac{1}{4}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right)$

(a) $e^2-3$ (b) $\frac{8}{3}$ (c) $\frac{9}{2}$

$\int_{x=0}^{x=1}\int_{y=x}^{y=2-x} f(x,y)\ \dee{y}\,\dee{x}$

(a)

(b) $\half(e-1)$

(a) The region $R$ is the shaded region in the figure

(b) $I= \int_{1/\sqrt{2}}^1 \int_{1/x}^{2x} f(x,y)\,\dee{y}\,\dee{x} +\int_1^2 \int_{x^2}^{2x} f(x,y)\,\dee{y}\,\dee{x}$ (c) $\frac{1}{2}$

(a) $\frac{32}{3}$

(b)

(c) $I = \int_0^1\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ x +\int_1^9\dee{y} \int_{(y-3)/2}^{\sqrt{y}}\dee{x}\ x$

The antiderivative of the function $\sin(y^2)$ cannot be expressed in terms of familiar functions. So we do not want the inside integral to be over $y$.

$\frac{3}{4}\big[1-\cos(16)\big]$

The inside integral, $\int_{\sqrt{y}}^1 \frac{\sin(\pi x^2)}{x}\ \dee{x}$, in the given form of $I$ looks really nasty. So try exchanging the order of integration.

(a)

(b) $\frac{1}{\pi}$

(a) $I=\int_0^1\dee{x}\int_x^1\dee{y}\ y^2\sin xy =\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy$

(b) $\frac{1-\sin 1}{2}$

$\frac{1-e^{-1}}{2}$

(a)

(b) $I=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x} +\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x}$ (c) $2\ln 2 -1$

The inside integral, $\int_{\sqrt{x}}^1 \sqrt{1+y^3}\ \dee{y}$, of the given integral looks pretty nasty. Try reversing the order of integration.

(a)

(b) $\frac{2\big(2\sqrt{2}-1\big)}{9}$

(b) The inside integral, $\int_{\frac{1}{2}\sqrt{x}}^1 e^{y^3}\ \dee{y}$, looks pretty nasty because $e^{y^3}$ does not have an obvious antiderivative. Try reversing the order of integration.

(a) $J=\frac{27}{4}$

(b) $I=\frac{4}{3}\big[e-1\big]$

(b) The inside integral, $\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x}$ looks nasty. Try reversing the order of integration.

(a)

(b) $\frac{\sin(8)}{3}$

(a) $I = \int_{-1}^2\int_{x^2}^{x+2} f(x,y)\ \dee{y}\,\dee{x}$ (b) $2e^2 + \frac{1}{e}$

$\frac{1}{9-x^2} =\frac{1}{6}\left(\frac{1}{x+3}-\frac{1}{x-3}\right)$.

(a)

(b) $\int_0^2\int_0^{x^2} f(x,y)\,\dee{y}\,\dee{x} +\int_2^{\sqrt{8}}\int_0^{8-x^2} f(x,y)\,\dee{y}\,\dee{x}$ (c) $\sqrt{8}-\arctan 2 -\frac{1}{6}\left[\ln\frac{3+\sqrt{8}}{3-\sqrt{8}} -\ln 5\right]$

The antiderivative of the function $e^{-y^2}$ cannot be expressed in terms of elementary functions. So the inside integral $\int_{-2}^{2x} e^{y^2}\ \dee{y}$ cannot be evaluated using standard calculus 2 techniques. Try reversing the order of integration.

$\frac{1}{4}\big[e^4-1\big]$

$I = \int_0^2 \int_y^{6-y^2} f(x,y)\ \dee{x}\,\dee{y}$

(a)

(b) $\int_0^1 \int_{-\sqrt{1-y}}^{\sqrt{1-y}} f(x,y)\ \dee{x}\,\dee{y}$, $\int_{-1}^1 \int_0^{1-x^2} f(x,y)\ \dee{y}\,\dee{x}$

(c) $e^{2/3}-e^{-2/3}$

The inside integral, over $y$, looks pretty nasty because $\sin(y^3)$ does not have an obvious antiderivative. So try reversing the order of integration.

(a)

(b) $\frac{1-\cos(1)}{12}$

(a)

(b) $27+18\ln\frac{3}{2}\approx 34.30$

The inside integral, $\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}$, in the given integral looks really nasty. So try exchanging the order of integration.

$\frac{4}{3}\sin 8\approx 1.319$

(a) $\dst I=\dblInt_D (8+2xy)\ \dee{x}\dee{y}$ where $D=\Set{(x,y)}{x^2+(y-1)^2\le 1}$

(b) $\dst I =\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ (8+2xy) =\int_{-1}^1 \dee{x}\int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\dee{y}\ (8+2xy)$

(c) $8\pi$

The inside integral, $\dst\int_{\sqrt{y}}^3\sin\big(\pi x^3\big)\ \dee{x}$, in the given integral looks really nasty. So try exchanging the order of integration.

$\frac{2}{3\pi}\approx 0.212$

(b) $\dst\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y}$ (c) $0$

$36$

$48\sqrt{2}\,\pi$