Final Test Office Hours

1 polar coordinate practice

problem:

\int_{0}^2 \int_{0}^\sqrt{ 2x-x^2 }7\sqrt{ x^2 + y^2 }dydx

this tells us that y is running from 0 up to the curve $y=\sqrt{2x-x^2}$

this is some circle of the form $y^2=2x-x^2$ iframe which is this circle

$x^2+y^2=2x$ thisis also known as $r=2\cos \theta$

lets graph this

Desmos

we are looking to fin the top region

we can try to solve it, but it turns out to b a mess like

$r^2{\sin }^2\theta +r^2{\cos }^2\theta =2r\cos \theta$

so lets do it the normal way

${\int }_0^{\frac{\pi }{2}}{\int }_0^{2\cos \theta }7r^{2}drd\theta$

2 - vector function problem

we are given

$r(t)=⟨3t^2,\ln y,y\ln y⟩k@(3,0,0)$

lets use t=1

so k is

$k=\frac{|{\vec{r}}^{\prime }(t)\times {\vec{r}}^{\prime \prime }(t)|}{|{\vec{r}}^{\prime }(t){|}^3}$

plugging this in

${\vec{r}}^{\prime }={⟨6t,\frac{1}{t},\ln t+1⟩|}_{t=1}=⟨6,1,1⟩$

${\vec{r}}^{\prime \prime }={⟨6,-\frac{1}{t^2},\frac{1}{t}⟩|}_{t=1}=⟨6,-1,1⟩$

$|{\vec{r}}^{\prime }|=\sqrt{36+1+1}=\sqrt{38}$

i & j & k \\ 6 & 1 & 1 \\ 6 & -1 & 1 \end{vmatrix} = \langle 2,0,-12 \rangle$$ $$| \vec{r}' \times \vec{r}'' | = \sqrt{ 4+144 } = \sqrt{ 148 } = 2\sqrt{ 37 }$$ $$k = \frac{2\sqrt{ 37 }}{38\sqrt{ 38 }}$$ $$\begin{vmatrix} 1 & 2 & 3 \\ 6 & 1 & 1 \\ 6 & -1 & 1 \end{vmatrix} = 1$$ ## chain order thing $$\int_{0}^8 \int _{x^2}^{64} f(x,y)dydx$$ step 1: sketch $\mathcal{D}$ > [!Info] Desmos > > <iframe src="https://www.desmos.com/calculator/v3xpwaw3om" width=600 height="400" ></iframe> we are trying to find the area above the x^2 curve and under the y=64 line so, $$\int_{0}^{64} \int_{0}^\sqrt{ y }f dxdy$$ $$\int_{0}^1 \int_{\arcsin y = x}^ \frac{\pi}{2} f(x,y) \ dx \ dy$$ so $$y = \sin x$$ > [!Info] Desmos > > <iframe src="https://www.desmos.com/calculator/pigfcalyyu" width=600 height="400" ></iframe> now we are tying to find the area before the line x=pi/2, and under the curve of y = sinx ## another one - converting to a polar integral $$\int_{0}^2 \int_{0}^{\sqrt{ 4-x^2 }=y}e^{-x^2 - y^2} dy dx$$ sketching $\mathcal{D}$ yields a circle centered at the origin, with a radius of 2. we are looking for the top right quadrant of the circle thus yieilds $$\int_{0}^ \frac{\pi}{2} \int_{0}^2 e^{-r^2} r \ dr \ d\theta$$ for I1: $$\int_{0}^2 re^{-r^2} \ dr$$ $$u = r^2 \ \ \ du = 2r \ dr $$ $$\frac{1}{2} \int_{0}^4 e^{-u}du$$ $$= \left. - \frac{1}{2} e^{-u}\right| _{0}^4$$ $$= - \frac{1}{2} e^{-4} + \frac{1}{2}$$ I2 $$\int_{0}^ \frac{\pi}{2} \frac{1}{2} - \frac{1}{2}e^{-4} d \theta$$ $$\frac{\pi}{2} \left( \frac{1}{2 }-\frac{1}{2}e^{-4} \right)$$ $$\frac{\pi}{4}\left( 1- \frac{1}{e^4} \right) = \frac{\pi}{4} \left( \frac{e^4-1}{e^4} \right)$$ ## another one for double integral $$\int \int_{\mathcal{D}}x dA$$ $\mathcal{D}$ is the region between $x^2 +y^2 = 16$ and $x^2 + y^2 = 4x$ > [!Info] Desmos > > <iframe src="https://www.desmos.com/calculator/uv9eul3ghk" width=600 height="400" ></iframe> method 1 to go about solving this: 1. $x^2 + y^2 - 4x = 0$ $$x^2 - 4x + y^2 = 0$$ $$(x-2)^2 - 4 +y^2 = 0$$ $$(x-2)^2 + y^2 =4 \ \ \ r = 4\cos \theta$$ 2. $r^2 = 4r\cos \theta$ $r = 4\cos \theta$ $$I_{1}: \ \ \int r^2 \cos \theta \ dr$$ $$ \left. \frac{r^3}{3}\cos \theta \right|_{4\cos \theta}^4$$ $$\frac{64}{3}\cos \theta - \frac{64}{3}\cos^4 \theta$$ $$$\frac{64}{3}\int_{0}^ \frac{\pi}{2} \cos \theta - \cos^4 \theta \ d\theta$$ $$\frac{64}{3} \int _{0}^ \frac{\pi}{2} \cos \theta - \frac{64}{3}\int_{0}^ \frac{\pi}{2} \cos^4 \theta \ d\theta$$ now we just simplify mechanically $$\int_{o}^ \frac{\pi}{2} \left( \frac{1-\cos 2\theta}{2} \right)^2 d\theta$$ $$\int_{0}^ \frac{\pi}{2} \frac{1}{4} ( 1-2\cos 2 \theta + \cos^2(2\theta))d\theta$$ $$ \frac{1}{2} \int_{0}^ \frac{\pi}{2} 1+2\cos(2\theta)+\left( \frac{1+cos (4\theta)}{2} \right)d\theta$$ $$\frac{1}{4} \int_{0}^ \frac{\pi}{2} \frac{3}{2} + 2\cos(2\theta)+\frac{\cos(4\theta)}{2} d\theta$$ $$\frac{1}{4 } \left( \frac{3}{2}\theta + \sin (2\theta)+ \frac{\sin(4\theta)}{8} \right) d\theta$$