3.1: Practice Set

\[ \begin{align*} \int_0^1 \int_0^\pi x\sin(xy)\, dy\, dx &=\int_0^1   -\cos(xy)  \Big|_0^\pi \, dx
\\[4pt] &= \int_0^1 (1-\cos(\pi x) )\,dx
\\[4pt] &= \left[ x-\dfrac{1}{\pi}\sin(\pi x) \right]\Bigg|_0^1
\\[4pt] &= 1 \end{align*} \]

\[ \begin{align*} \int_0^1 \int_0^\pi y\sin(xy) \, dy\, dx &=\int_0^1 \left( \left[ y\cdot \left(-\dfrac{1}{x}\cos(xy)\right)\right]\Bigg|_0^\pi -\int_0^\pi -\dfrac{1}{x}\cos(xy) \,dy \right) \, dx
\\[4pt] &= \int_0^1 \left[ y\cdot \left(-\dfrac{1}{x}\cos(xy)\right)+\dfrac{1}{x^2}\sin(xy)\right]\Bigg|_0^\pi \,dx
\\[4pt] &= \int_0^1 \left( \dfrac{1}{x^2}\sin(\pi x)-\dfrac{\pi}{x}\cos(\pi x) \right)\,dx \end{align*} \]

This integral looks difficult! Instead we switch the order of integration (which is easy since we are integrating over a rectangular domain):

\[ \begin{align*} \int_0^1 \int_0^\pi y\sin(xy) \, dy\, dx &= \int_0^\pi \int_0^1 y\sin(xy) \, dx\, dy
\\[4pt] &= \int_0^\pi   -\cos(xy)  \Big|_0^1 \, dy
\\[4pt] &= \int_0^\pi \left(1-\cos(y) \right)\,dy
\\[4pt] &= \left[ y-\sin(y) \right]\Big|_0^\pi
\\[4pt] &= \pi \end{align*} \]

\[ \begin{align*} \iint_D \dfrac{12}{y+yx^2}\, dA &= \int_1^{\sqrt{3}}\int_1^e \dfrac{12}{y(1+x^2)} \,dy \, dx
\\[4pt] &= \left( \int_1^{\sqrt{3}} \dfrac{12}{1+x^2}\, dx \right) \left( \int_1^e \dfrac{1}{y} \, dy \right)
\\[4pt] &= \left( 12\tan^{-1}(x) \Big|_1^{\sqrt{3}}\right) \cdot \left( \ln|y| \Big|_1^e\right)
\\[4pt] &= \left( 4\pi -3\pi \right)\cdot \left( \ln(e)-\ln(1)\right)
\\[4pt] &= \pi \end{align*} \]

\[ \begin{align*} \int_0^1 \int_0^x \cos(\pi x^2)\, dy\, dx&= \int_0^1 y\cos(\pi x^2)  \Big|_0^x \, dx
\\[4pt] &= \int_0^1 x\cos(\pi x^2) \, dx
\\[4pt] &=   \dfrac{1}{2\pi} \sin(\pi x^2) \Bigg|_0^1
\\[4pt] &= \dfrac{1}{2\pi}\sin(\pi)-\dfrac{1}{2\pi}\sin(0)
\\[4pt] &=0 \end{align*} \]

\[ \begin{align*} \int_1^2 \int_1^{x^3} \frac{x^4}{y^3} \, dy \, dx &= \int_1^2  \dfrac{1}{-2} x^4y^{-2} \Bigg|_1^{x^3} \, dx
\\[4pt] &= \int_1^2 \left(-\dfrac{1}{2}x^4\cdot x^{-6}+\dfrac{1}{2}x^4\cdot 1^{-2} \right) \, dx
\\[4pt] &= \int_1^2 \left(-\dfrac{1}{2} x^{-2}+\dfrac{1}{2}x^4\right) \, dx
\\[4pt] &= \left[ \dfrac{1}{2}x^{-1}+\dfrac{1}{10}x^5 \right]\Bigg|_1^2
\\[4pt] &= \left( \dfrac{1}{4}+\dfrac{16}{5}\right)-\left( \dfrac{1}{2}+\dfrac{1}{10}\right)
\\[4pt] &= \dfrac{57}{20} \end{align*} \]

This integral \(dx\) is not simple to calculate, so we instead change the order of integration. First we sketch the region we are integrating over:

We can write it as \(D=\left\{ (x,y)\, : \, 0\le y\le \sqrt{3},\, 0\le x\le 3-y^2\right\} = \left\{ (x,y)\, : \, 0\le x\le 3, \, 0\le y\le \sqrt{3-x}\right\}\), so

\[ \begin{align*} \int_0^{\sqrt{3}} \int_0^{3-y^2} 16y\cdot \sqrt[3]{ x^2-6x+1} \, dx\, dy &= \int_0^3 \int_0^{\sqrt{3-x}}16y\cdot \sqrt[3]{ x^2-6x+1} \, dy \, dx
\\[4pt] &= \int_0^3   8y^2\cdot \sqrt[3]{ x^2-6x+1}   \Big|_0^{\sqrt{3-x}} \, dx
\\[4pt] &= \int_0^3 \left(8 (3-x) \cdot \sqrt[3]{ x^2-6x+1}\right) \, dx
\\[4pt] &=   -3(x^2-6x+1)^{\frac{4}{3}}  \Big|_0^3
\\[4pt] &= -3\left((-8)^{\frac{4}{3}}-1\right)
\\[4pt] &= -45 \end{align*} \]

This integral \(dy\) is not simple to calculate, so we instead change the order of integration. First we sketch the region we are integrating over:

We can write it as \(D=\left\{ (x,y)\, : \, 0\le x\le \dfrac{1}{2},\, 2x \le y\le 1\right\} = \left\{ (x,y)\, : \, 0\le y\le 1, \, 0\le x\le \dfrac{y}{2}\right\}\), so

\[ \begin{align*} \int_0^{\frac{1}{2}} \int_{2x}^{1} \dfrac{1}{\sqrt{1-y^4}}\, dy\, dx &= \int_0^1 \int_0^{\frac{y}{2}} \dfrac{1}{\sqrt{1-y^4}} \, dx \, dy
\\[4pt] &= \int_0^1 \dfrac{1}{\sqrt{1-y^4}}x \Bigg|_0^{\frac{y}{2}} \, dy
\\[4pt] &= \int_0^1 \dfrac{y}{2\sqrt{1-y^4}} \, dy
\\[4pt] &=   \dfrac{1}{4}\sin^{-1}(y^2) \Bigg|_0^1
\\[4pt] &= \dfrac{1}{4}\left( \sin^{-1}(1)-\sin^{-1}(0)\right)
\\[4pt] &= \dfrac{\pi}{8} \end{align*} \]

\[ \begin{align*} \int_\pi^{2\pi} \int_y^{2y} \sin(y)\sin^3(x)\cos(x)\, dx\, dy &= \int_\pi^{2\pi}  \dfrac{1}{4}\sin(y)\sin^4(x) \Bigg|_y^{2y} \, dy
\\[4pt] &= \dfrac{1}{4} \int_\pi^{2\pi} \left( \sin(y)\sin^4(2y)-\sin^5(y) \right) \, dy
\\[4pt] &= \dfrac{1}{4} \int_\pi^{2\pi} \left( 16\sin^5(y)\cos^4(y)-\sin^5(y) \right) \, dy
\\[4pt] &= \dfrac{1}{4} \int_\pi^{2\pi} ( 16\cos^4(y)-1) (1-\cos^2(y))^2 \sin(y) \, dy
\\[4pt] &= \dfrac{1}{4} \int_{-1}^1 (16w^4-1) (1-w^2)^2 \, (-dw)
\\[4pt] &= -\dfrac{1}{4} \int_{-1}^1 (16w^8-32w^6+15w^4+2w^2-1) \, dw
\\[4pt] &= -\dfrac{1}{4}\left[ \dfrac{16}{9}w^9 -\dfrac{32}{7}w^7+3w^5+\dfrac{2}{3}w^3-w \right]\Bigg|_{-1}^1
\\[4pt] &= -\dfrac{1}{4}\left( \dfrac{32}{9}-\dfrac{64}{7}+6+\dfrac{4}{3}-2\right)
\\[4pt] &= \dfrac{4}{63} \end{align*} \]

We see the domain we are integrating over is a triangle:

We can define this triangle most easily as a Type 1 region: \(T=\left\{ (x,y)\, : \, 0\le x\le 2,\, -2x\le y \le x \right\} \), so

\[ \begin{align*} \iint_T 5xy^2\, dA &= \int_0^2 \int_{-2x}^x 5xy^2 \, dy \, dx
\\[4pt] &= \int_0^2 \dfrac{5}{3} xy^3 \Bigg|_{-2x}^x\, dx
\\[4pt] &= \int_0^2 15x^4 \, dx
\\[4pt] &=   3x^5   \Big|_0^2
\\[4pt] &= 96 \end{align*} \]

\[ \begin{align*} \int_1^2 \int_{x+1}^{2x+1} \dfrac{1}{y^3} \, dy \, dx &= \int_1^2   -\dfrac{1}{2}y^{-2} \Bigg|_{x+1}^{2x+1} \, dx
\\[4pt] &= -\dfrac{1}{2} \int_1^2 \left( (2x+1)^{-2}-(x+1)^{-2} \right) \, dx
\\[4pt] &= -\dfrac{1}{2} \left[ -\dfrac{1}{2}(2x+1)^{-1}+(x+1)^{-1} \right]\Bigg|_1^2
\\[4pt] &= -\dfrac{1}{2}\left( -\dfrac{1}{10}+\dfrac{1}{3}+\dfrac{1}{6}-\dfrac{1}{2}\right)
\\[4pt] &= \dfrac{1}{20} \end{align*} \]

\[ \begin{align*} \int_0^1 \int_{x^2}^x xe^y \, dy \, dx &= \int_0^1  xe^y   \Big|_{x^2}^x \, dx
\\[4pt] &= \int_0^1 \left( xe^x-xe^{x^2}\right) \, dx
\\[4pt] &= \left[ (x-1)e^x-\dfrac{1}{2}e^{x^2}\right]\Bigg|_0^1
\\[4pt] &= \dfrac{3-e}{2} \end{align*} \]

\[ \begin{align*} \int_3^4 \int_0^6 \frac{y}{x^3-x^2-2x}\, dy \, dx &= \int_3^4   \dfrac{1}{2x(x-2)(x+1)}y^2 \Bigg|_{0}^6 \, dx
\\[4pt] &= \int_3^4 \dfrac{18}{x(x-2)(x+1)} \, dx
\\[4pt] &= \int_3^4 \left( -\dfrac{9}{x}+\dfrac{3}{x-2}+\dfrac{6}{x+1} \right) \, dx
\\[4pt] &= \left[ -9\ln|x|+3\ln|x-2|+6\ln|x+1| \right] \Big|_3^4
\\[4pt] &= \left( -9\ln(4)+3\ln(2) +6\ln (5) \right) - \left( -9\ln(3)+3\ln(1)+6\ln(4)\right)
\\[4pt] &= 6\ln(5)+9\ln(3)-27\ln(2) \end{align*} \]

We see the domain we are integrating over is a trapezoid:

This is already given as a Type 1 region, so:

\[ \begin{align*} \iint_D \dfrac{1}{(x+y)^2} \,dA &= \int_1^2 \int_0^x (x+y)^{-2} \, dy\, dx
\\[4pt] &= \int_1^2   -(x+y)^{-1}  \Big|_0^x \, dx
\\[4pt] &= \int_1^2 \left( -\dfrac{1}{2x}+\dfrac{1}{x} \right) \, dx
\\[4pt] &= \int_1^2 \dfrac{1}{2x} \, dx
\\[4pt] &=  \dfrac{1}{2}\ln|x| \Bigg|_1^2
\\[4pt] &= \dfrac{1}{2}\ln(2)-\dfrac{1}{2}\ln(1)
\\[4pt] &= \dfrac{1}{2}\ln(2) \end{align*} \]

\[ \begin{align*} \int_1^4 \int_{0}^{3} 9\sqrt{x+xy} \, dy\, dx &= \int_1^4 \int_{0}^{3} 9\sqrt{x}\cdot \sqrt{1+y} \, dy \, dx
\\[4pt] &= \left( \int_1^4 9\sqrt{x}\,dx \right) \left( \int_0^3 \sqrt{1+y}\, dy \right)
\\[4pt] &= \left( 6x^{\frac{3}{2}}  \Big|_1^4\right) \cdot \left( \dfrac{2}{3}(1+y)^{\frac{3}{2}} \Bigg|_0^3\right)
\\[4pt] &= 42 \cdot \dfrac{14}{3}
\\[4pt] &= 196 \end{align*} \]

\[ \begin{align*} \int_0^\pi \int_{\cos(x)}^1 y^8\sin(x) \, dy\, dx &= \int_0^\pi  \dfrac{1}{9}y^9\sin(x) \Bigg|_{\cos(x)}^1 \, dx
\\[4pt] &= \dfrac{1}{9} \int_0^\pi \left(1-\cos^9(x)\right)\sin(x) \, dx
\\[4pt] &= \dfrac{1}{9} \int_1^{-1} \left(1-w^9\right) \, (-dw)
\\[4pt] &= \dfrac{1}{9} \int_1^{-1} \left(w^9-1\right) \, dw
\\[4pt] &= \dfrac{1}{9} \left[ \dfrac{1}{10}w^{10}-w \right]\Bigg|_1^{-1}
\\[4pt] &= \dfrac{2}{9} \end{align*} \]

We sketch the region \(E\) with given edges:

This is best described as a Type 2 region \(E=\left\{ (x,y)\, : \, -\dfrac{\pi}{2}\le y \le \dfrac{\pi}{2},\, \sin(y)\le x\le 1 \right\}\), so:

\[ \begin{align*} \iint_E x^3 \,dA &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{\sin(y)}^1 x^3\, dx \, dy
\\[4pt] &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}   \dfrac{1}{4}x^4 \Bigg|_{\sin(y)}^1 \, dy
\\[4pt] &= \dfrac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( 1-\sin^4(y)\right) \, dy
\\[4pt] &= \dfrac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(-\dfrac{1}{8}\cos(4y)+\dfrac{1}{2}\cos(2y)+\dfrac{5}{8}\right) \, dy
\\[4pt] &= \dfrac{1}{4} \left[-\dfrac{1}{32}\sin(4y)+\dfrac{1}{4}\sin(2y)+\dfrac{5}{8}y \right]\Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\\[4pt] &= \dfrac{1}{4} \cdot \dfrac{5\pi}{8}
\\[4pt] &= \dfrac{5\pi}{32} \end{align*} \]