[11.2] - Double Integrals - Practice

1. $\int_0^2\int_0^2(x^2-y^2)\,dy\,dx$
   
   
   
   
   
   
2. $\int_0^{\frac{\pi}{4}}\int_{y=0}^{\frac{\pi}{2}}\cos(2x+y)\,dy\,dx$
   Hint: According to one of the Trig addition formulas, $\cos(\alpha+\beta)= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$
   
   
   
   
   
   
3. $\int_0^2\int_{y=1}^3x^3y\,dy\,dx$
   
   
   
   
   
   
4. $\int_{-1}^{1}\int_{y=0}^{\pi}x^2\sin y\,dy\,dx$
   
   $$\begineq \iint &=\int_{-1}^{1}{\color{blue}\int_{y=0}^{\pi}x^2\sin y\,dy}\,dx\\ &=\int_{-1}^{1}{\color{blue}x^2\int_{y=0}^{\pi}\sin y\,dy}\,dx =\int_{-1}^{1}{\color{blue}x^2\Big(-\cos y\Big)_0^{y=\pi}} \,dx\\ &=\int_{-1}^{1}{\color{blue}x^2\left(-(-1) - (-(1))\right)}\,dx \\ &={\color{blue}2}\left(\frac13x^3\right)_{-1}^{x=+1}\,dx = {\color{red}\frac43}.\\ \endeq $$
   
   
   
   
   
   
5. $\int_0^2\int_{y=x^2}^{2x}(x^2+2y)\,dy\,dx$
   
   $$\begineq \iint &= \int_0^2{\color{blue}\int_{y=x^2}^{2x}(x^2+2y)\,dy}\,dx \\ &= \int_0^2{\color{blue}\Big(x^2y+y^2\Big)^{y=2x}_{x^2}}\,dx\\ &= \int_0^2{\color{blue}\Big(x^2\cdot 2x+(2x)^2-x^2\cdot x^2-(x^2)^2\Big)} \,dx \\ &= \int_0^2{\color{blue}\Big(2x^3+4x^2-2x^4\Big)} \,dx \\ &= \Big(\frac24x^4+\frac43x^3-\frac25x^5\Big)_0^{x=2} \\ &= \frac12(16)\frac{15}{15}+\frac43(8)\frac55-\frac25(32)\frac33 =\frac{120}{15}+\frac{160}{15}-\frac{192}{15}\\ \\&=\color{red}\frac{88}{15}\approx 5.867 \endeq $$
   
   
   
   
   
   
6. $\int_0^3\int_{y=0}^{9-x^2}4x\,dy\,dx$