Question

Evaluate the following integral over the Region R. (Answer accurate to 2 decimal places). $\iint_R 1(x+y) dA$ $R = \{(x, y) | 1 \le x^2 + y^2 \le 36, x \le 0\}$ Hint: Be sure to convert to Polar coordinates. Use $dA = r dr d\theta$.

          Evaluate the following integral over the Region R.
(Answer accurate to 2 decimal places).
$\iint_R 1(x+y) dA$
$R = \{(x, y) | 1 \le x^2 + y^2 \le 36, x \le 0\}$
Hint: Be sure to convert to Polar coordinates. Use $dA = r dr d\theta$.

$Evaluate the following integral over the Region R. (Answer accurate to 2 decimal places). 1(x+y) dA R = {(x, y) | 1 ≤ x^2 + y^2 ≤ 36, x ≤ 0} Hint: Be sure to convert to Polar coordinates. Use dA = r dr dθ.$

Added by Melissa A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Likhit Ganedi

07/13/2023

Step 1

$$ x = r cos(\theta) $$ $$ y = r sin(\theta) $$ $$ dA = r dr d\theta $$ $$ 1 \le x^2 + y^2 \le 36 $$ $$ 1 \le r^2 \le 36 $$ $$ 1 \le r \le 6 $$ $$ x \le 0 $$ $$ r cos(\theta) \le 0 $$ $$ cos(\theta) \le 0 $$ $$ \frac{\pi}{2} \le \theta \le \frac{3\pi}{2} $$ $$ Show more…

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Evaluate the following integral over the Region R. (Answer accurate to 2 decimal places). $$ \int \int_R 1(x+y) dA $$ $$ R = \{(x, y) | 1 \le x^2 + y^2 \le 36, x \le 0 \} $$ Hint: Be sure to convert to Polar coordinates. Use dA = r dr d $\theta$.