Answered: 1 5-y3 x + y dædy

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.3: Quadratic Equations

Problem 51E

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Question

Convert the integral into polar coordinates. (do not evaluate).

Expert Solution

Step 1

$T h e g i v e n i n t e g r a l \int_{0}^{1} \int_{y}^{\sqrt{5 - y^{3}}} (x + y) d x d y i s i n t h e c a r t e s i a n c o - o r d i n a t e w h e r e y \leq x \leq \sqrt{5 - y^{3}} a n d 0 \leq y \leq 1 w e h a v e t o c o n v e r t i t i n t o p o l a r c o o r d i n a t e s . f o r t h i s, c o n s i d e r x = r \cos θ y = r \sin θ r^{2} = x^{2} + y^{2} \Rightarrow r = \sqrt{x^{2} + y^{2}} a n d \tan θ = \frac{y}{x}$

Step 2

$T h e J a c o b i a n d e t e r m i n a n t f o r t h i s t r a n s f o r m a t i o n i s \frac{\partial (x, y)}{\partial (r, θ)} = |\begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{matrix}| = |\begin{matrix} \frac{\partial (r \cos θ)}{\partial r} & \frac{\partial (r \cos θ)}{\partial θ} \\ \frac{\partial (r \sin θ)}{\partial r} & \frac{\partial (r \sin θ)}{\partial θ} \end{matrix}| = |\begin{matrix} \cos θ & - r \sin θ \\ \sin θ & r \cos θ \end{matrix}| = \cos θ \cdot r \cos θ - (- r \sin θ) \cdot \sin θ = r \cos^{2} θ + r \sin^{2} θ = r (\cos^{2} θ + \sin^{2} θ) = r [\cos^{2} θ + \sin^{2} θ = 1] d x d y = |\frac{\partial (x, y)}{\partial (r, θ)}| d r d θ = r d r d θ$

Step 3

$s o w e c a n w r i t e \iint_{D} f (x, y) d x d y = \int_{θ = α}^{θ = β} \int_{r = a}^{r = b} f (r \cos θ, r \sin θ) r d r d θ = \int_{θ = α}^{θ = β} \int_{r = a}^{r = b} f (r, θ) r d r d θ w h e r e D i s t h e g i v e n r e g i o n n o w f (x, y) = x + y \Rightarrow f (r, θ) = r \cos θ + r \sin θ = r (\cos θ + \sin θ) N o w w e h a v e t o c h a n g e t h e l i m i t s o f i n t e g r a l s .$

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