Question-126632

Question Number 126632 by BHOOPENDRA last updated on 22/Dec/20

Commented by BHOOPENDRA last updated on 22/Dec/20

t h a n k u s i r

Answered by Ar Brandon last updated on 22/Dec/20

In polar coordinates;

{\begin{cases} x = rcos θ & r ⩾ 0 \\ y = rsin θ & θ \in [0, 2 π] \end{cases}

1 ⩽ x^{2} + y^{2} ⩽ 4 \Rightarrow 1 ⩽ r^{2} ⩽ 4 \Rightarrow 1 ⩽ r ⩽ 2

x ⩽ 0 \Rightarrow rcos θ ⩽ 0 \Rightarrow \cos θ ⩽ 0 \Rightarrow θ \in [\frac{π}{2}, \frac{3 π}{2}]

I = \int \int_{R} (x + y) dA

= \int_{\frac{3 π}{2}}^{\frac{π}{2}} \int_{1}^{2} r (rcos θ + rsin θ) drd θ

= \int_{\frac{3 π}{2}}^{\frac{π}{2}} \int_{1}^{2} r^{2} (\cos θ + \sin θ) drd θ

= {[\frac{r^{3}}{3}]}_{1}^{2} {[\sin θ - \cos θ]}_{\frac{3 π}{2}}^{\frac{π}{2}}

= (\frac{8}{3} - \frac{1}{3}) (1 - - 1) = \frac{14}{3}

Answered by mathmax by abdo last updated on 22/Dec/20

A = \int \int_{R} (x + y) dxdy with R = {(x, y) / 1 ⩽ x^{2} + y^{2} ⩽ 4 and x ⩽ 0}

we use the diffeomorphism {\begin{cases} x = rcos θ \\ y = rsin θ \end{cases}

1 ⩽ x^{2} + y^{2} ⩽ 4 \Rightarrow 1 ⩽ r^{2} ⩽ 4 \Rightarrow 1 ⩽ r ⩽ 2

x ⩽ 0 \Rightarrow \cos θ ⩽ 0 \Rightarrow θ \in [\frac{π}{2}, \frac{3 π}{2}] \Rightarrow

A = \int_{1}^{2} \int_{\frac{π}{2}}^{\frac{3 π}{2}} (rcos θ + rsin θ) rdrd θ = \int_{1}^{2} r^{2} dr \int_{\frac{π}{2}}^{\frac{3 π}{2}} (\cos θ + \sin θ) d θ

= {[\frac{r^{3}}{3}]}_{1}^{2} \times {[\sin θ - \cos θ]}_{\frac{π}{2}}^{\frac{3 π}{2}} = \frac{1}{3} (8 - 1) (\sin (\frac{3 π}{2}) - \cos (\frac{3 π}{2}) - \sin (\frac{π}{2}) + \cos (\frac{π}{2}))

= \frac{7}{3} (- 1 - o - 1) = - \frac{14}{3}