Question

(c) \iint_D (a - \sqrt{x^2 + y^2})dA, where D = \{(x, y) : x^2 + y^2 \le a^2\}. (d) \iint_D \sqrt{b^2 - y^2}dA, where D = \{(x, y) : 0 \le x \le a, 0 \le y \le b\}.

          (c) \iint_D (a - \sqrt{x^2 + y^2})dA, where D = \{(x, y) : x^2 + y^2 \le a^2\}.
(d) \iint_D \sqrt{b^2 - y^2}dA, where D = \{(x, y) : 0 \le x \le a, 0 \le y \le b\}.

$(c) (a - √(x^2 + y^2))dA, where D = {(x, y) : x^2 + y^2 ≤a^2}. (d) √(b^2 - y^2)dA, where D = {(x, y) : 0 ≤x ≤a, 0 ≤y ≤b}.$

Added by Nicholas Z.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

12/15/2023

Step 1

We can use polar coordinates: $x = r \cos \theta$, $y = r \sin \theta$, and $dA = r dr d\theta$. The region $D$ is described by $0 \le r \le a$ and $0 \le \theta \le 2\pi$. The integral becomes: $$ \int_0^{2\pi} \int_0^a (a - r) r dr d\theta $$ Show more…

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(c) $\iint_D (a - \sqrt{x^2 + y^2}) dA$, where $D = \{(x, y) : x^2 + y^2 \le a^2\}$. (d) $\iint_D \sqrt{b^2 - y^2} dA$, where $D = \{(x, y) : 0 \le x \le a, 0 \le y \le b\}$.