Question

S (x+2y2) dx dy =? 0 ſv 1 cos s (psino)dpdo=? 0 0

Answer 1

Solution a) So' S (x +202) dx dy - 1 **+2'de [ (x) = g(x) dx = [ {(x) dx = / 3 (x) dx - Sca [ 27?dx - xx = (1-1 Apply the Sum Rule : xdx + Compute the boundaries : SI f(x) dx = F (b) – F(a) lim (F(x)) - lim (F(x)) x-b- Xat a X V (VT) lim x++ (8) Plug in the value x = 2 2

$\\\mathrm{\Rightarrow \lim _{x\to \:1-}\left(\frac{x^2}{2}\right)\mathrm{Plug\:in\:the\:value}\:x=1=\frac{1^2}{2}=\frac{1}{2}}\\ \\ \mathrm{=\frac{1}{2}-\frac{y}{2}}\\ \\ \mathrm{\Rightarrow \int _{\sqrt{y}}^12y^2dx=\left[2y^2x\right]^1_{\sqrt{y}}}\\ \\ \mathrm{\mathrm{Compute\:the\:boundaries}:\quad \left[2y^2x\right]^1_{\sqrt{y}}}\\ \\ \mathrm{\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)}\\ \\ \mathrm{\Rightarrow \lim _{x\to \sqrt{y}+}\left(2y^2x\right)\mathrm{Plug\:in\:the\:value}\:x=\sqrt{y}=2y^2\sqrt{y}}\\ \\ \mathrm{\Rightarrow \lim _{x\to \:1-}\left(2y^2x\right)\mathrm{Plug\:in\:the\:value}\:x=1=2y^2\cdot \:1=2y^2}\\ \\ \mathrm{=2y^2-2y^2\sqrt{y}}\\ \\ \mathrm{=\frac{1}{2}-\frac{y}{2}+2y^2-2y^2\sqrt{y}}\\ \\ \mathrm{=\int _0^1\left(\frac{1}{2}-\frac{y}{2}+2y^2-2y^2\sqrt{y}\right)dy}\\ \\$

$\\\mathrm{\Rightarrow }\mathrm{\mathrm{\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}}\\ \\ \mathrm{=\int _0^1\frac{1}{2}dy-\int _0^1\frac{y}{2}dy+\int _0^12y^2dy-\int _0^12y^2\sqrt{y}dy}\\ \\ \mathrm{\Rightarrow \int _0^1\frac{1}{2}dy=\left[\frac{1}{2}y\right]_0^1=\frac{1}{2}}\\ \\ \mathrm{\Rightarrow \int _0^1\frac{y}{2}dy=\frac{1}{2}\cdot \int _0^1ydy=\frac{1}{2}\left[\frac{y^{1+1}}{1+1}\right]_0^1=\frac{1}{2}\left[\frac{y^2}{2}\right]_0^1=\frac{1}{2}=\frac{1}{2}\cdot \frac{1}{2}=4}\\ \\ \mathrm{\Rightarrow \int _0^12y^2dy=2\cdot \int _0^1y^2dy=2\left[\frac{y^{2+1}}{2+1}\right]_0^1=2\left[\frac{y^3}{3}\right]_0^1=\frac{1}{3}=\frac{2}{3}}\\ \\ \mathrm{\Rightarrow \int _0^12y^2\sqrt{y}dy=2\cdot \int _0^1y^2\sqrt{y}dy=2\cdot \int _0^1y^{\frac{5}{2}}dy=2\left[\frac{y^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_0^1=2\left[\frac{2}{7}y^{\frac{7}{2}}\right]_0^1=\frac{2}{7}=2\cdot \frac{2}{7}=\frac{4}{7}}\\ \\$$\\\mathrm{=\frac{1}{2}-\frac{1}{4}+\frac{2}{3}-\frac{4}{7}}\\ \\ \mathrm{=\frac{29}{84}}$

$\\\mathrm{b)}\\ \\ \mathrm{\int _0^{\pi }\:\int _0^{cos\theta }\:\left(\rho \:sin\theta \right)d\rho \:d\theta }\\ \\ \mathrm{\Rightarrow \int _0^{\cos \left(\theta \right)}\rho \sin \left(\theta \right)d\rho }\\ \\ \mathrm{=\sin \:\left(\theta \right)\cdot \:\int _0^{\cos \:\left(\theta \right)}\rho d\rho =\sin \:\left(\theta \right)\left[\frac{\rho ^{1+1}}{1+1}\right]^{\cos \:\left(\theta \right)}_0=\sin \:\left(\theta \right)\left[\frac{\rho ^2}{2}\right]^{\cos \:\left(\theta \right)}_0}\\ \\ \mathrm{\mathrm{Compute\:the\:boundaries}:\quad \:\left[\frac{\rho ^2}{2}\right]^{\cos \:\left(\theta \right)}_0}\\ \\ \mathrm{\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)}\\ \\ \mathrm{\Rightarrow \lim _{\rho \to \:0+}\left(\frac{\rho ^2}{2}\right)}\\ \\ \mathrm{\mathrm{Plug\:in\:the\:value}\:\rho =0}\\ \\ \mathrm{=\frac{0^2}{2}=0}\\ \\$

$\\\mathrm{\lim _{\rho \to \cos \left(\theta \right)-}\left(\frac{\rho ^2}{2}\right)}\\ \\ \mathrm{\mathrm{Plug\:in\:the\:value}\:\rho =\cos \left(\theta \right)}\\ \\ \mathrm{=\frac{\cos ^2\left(\theta \right)}{2}}\\ \\ \mathrm{=\frac{\cos ^2\left(\theta \right)}{2}-0=\frac{\cos ^2\left(\theta \right)}{2}}\\ \\ \mathrm{=\sin \left(\theta \right)\frac{\cos ^2\left(\theta \right)}{2}}\\ \\ \mathrm{\Rightarrow \int _0^{\pi }\sin \left(\theta \right)\frac{\cos ^2\left(\theta \right)}{2}d\theta }\\ \\$

$\\\mathrm{=\frac{1}{2}\cdot \:\int _0^{\pi \:}\sin \:\left(\theta \right)\cos \:^2\left(\theta \right)d\theta }\\ \\ \mathrm{\mathrm{Apply\:u-substitution:}\:u=\cos \left(\theta \right)}\\ \\ \mathrm{\frac{du}{d\theta }=-\sin \:\left(\theta \right)\Rightarrow \:\:du=-\sin \:\left(\theta \right)d\theta \Rightarrow \:\:d\theta =\left(-\frac{1}{\sin \:\left(\theta \right)}\right)du}\\ \\ \mathrm{=\int \:\sin \:\left(\theta \right)u^2\left(-\frac{1}{\sin \:\left(\theta \right)}\right)du=\int \:\:-u^2du}\\ \\ \mathrm{Adjust\:integral\:boundaries:}\\ \\ \mathrm{\theta =0\quad \Rightarrow \:u=1}\\ \\ \mathrm{\theta =\pi \quad \Rightarrow \:u=-1}\\ \\ \mathrm{=\int _1^{-1}-u^2du}\\ \\ \mathrm{=\frac{1}{2}\cdot \int _1^{-1}-u^2du}\\ \\ \mathrm{=\frac{1}{2}\cdot \int _1^{-1}-u^2du}\\ \\ \mathrm{\int _a^bf\left(x\right)dx=-\int _b^af\left(x\right)dx,\:a<b}\\ \\ \mathrm{=\frac{1}{2}\left(-\int _{-1}^1-u^2du\right)}\\ \\ \mathrm{=\frac{1}{2}\left(-\left(-\int _{-1}^1u^2du\right)\right)}\\ \\$

$\\\mathrm{=\frac{1}{2}\left(-\left(-\left[\frac{u^{2+1}}{2+1}\right]^1_{-1}\right)\right)}\\ \\ \mathrm{=\frac{1}{2}\left[\frac{u^3}{3}\right]^1_{-1}}\\ \\ \mathrm{\mathrm{Compute\:the\:boundaries}:\quad \left[\frac{u^3}{3}\right]^1_{-1}}\\ \\ \mathrm{\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)}\\ \\ \mathrm{\Rightarrow \lim _{u\to \:-1+}\left(\frac{u^3}{3}\right)\mathrm{Plug\:in\:the\:value}\:u=-1=\frac{\left(-1\right)^3}{3}=-\frac{1}{3}}\\ \\ \mathrm{\Rightarrow\lim _{u\to \:1-}\left(\frac{u^3}{3}\right) \mathrm{Plug\:in\:the\:value}\:u=1=\frac{1^3}{3}=\frac{1}{3}}\\ \\ \mathrm{=\frac{1}{3}-\left(-\frac{1}{3}\right)}\\ \\ \mathrm{=\frac{2}{3}}\\ \\ \mathrm{=\frac{1}{2}\cdot \frac{2}{3}}\\ \\ \mathrm{=\frac{1}{3}}\\ \\$