The integral 1-2- Dx-y-10dx dy- where D denotes the disc- x2-y2- 4- evaluates to

Let x=r cos θ, y=r sin θ, then 0 ≤ r ≤ 2 & 0 ≤θ≤ 2π & J=r I=12π∫∫_D [ (x+y)+10 ]dx dy =12π∫_r=0^2∫_θ =0^2π [ r(cos θ+sin θ)+10 ]r dr dθ =12π∫_r=0^2 ...

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The integral $\frac{1}{2 π} \int \int_{D} (x + y + 10) d x d y,$ where $D$ denotes the disc: $x^{2} + y^{2} \leq 4,$ evaluates to
20

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Solution

The correct option is A 20
Let $x = r c o s θ,$
$y = r s i n θ,$
then $0 \leq r \leq 2 &$
$0 \leq θ \leq 2 π & J = r$

$I = \frac{1}{2 π} \int \int_{D} [(x + y) + 10] d x d y$
$= \frac{1}{2 π} \int_{r = 0}^{2} \int_{θ = 0}^{2 π} [r (c o s θ + s i n θ) + 10] r d r d θ$
$= \frac{1}{2 π} \int_{r = 0}^{2} {[r (s i n θ - c o s θ) + 10 θ]}_{0}^{2 π} r . d r$
$= \frac{1}{2 π} (20 π {(\frac{r^{2}}{2})}_{0}^{2}) = 20$

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The integral 12π∫∫D(x+y+10)dxdy, where D denotes the disc: x2+y2≤4, evaluates to 20

The correct option is A 20Let x=rcosθ, y=rsinθ, then 0≤r≤2& 0≤θ≤2π&J=r I=12π∫∫D[(x+y)+10]dxdy =12π∫2r=0∫2πθ=0[r(cosθ+sinθ)+10]rdrdθ =12π∫2r=0[r(sinθ−cosθ)+10θ]2π0r.dr =12π(20π(r22)20)=20

The integral $\frac{1}{2 π} \int \int_{D} (x + y + 10) d x d y,$ where $D$ denotes the disc: $x^{2} + y^{2} \leq 4,$ evaluates to
20