Answered: "Using polar coordinates, evaluate the integral ∫∫Rsin⁡(x2+y2)dA where..."

Arthur Pratt

Answered question

2021-12-19

Using polar coordinates, evaluate the integral

\int \int_{R} \sin (x^{2} + y^{2}) d A

where R is region

4 \leq x^{2} + y^{2} \leq 64

Answer & Explanation

Hattie Schaeffer

Beginner2021-12-20Added 37 answers

Use the table of integrals to solve this problem

SlabydouluS62

Skilled2021-12-21Added 52 answers

But, I need explanation, please.

nick1337

Expert2021-12-28Added 777 answers

Using polar coordinates, we have $x = r \cos θ, y = r \sin θ$
$\Rightarrow x^{2} + y^{2} = r^{2}, θ = \tan^{- 1} (\frac{y}{x}), d A = r d r d θ$
Given $R = {(x, y) | 4 \leq x^{2} + y^{2} \leq 64} = {(r, θ) | 0 \leq θ \leq 2 x, 2 \leq r \leq 8}$
$\begin{matrix} \int \int_{R} \sin (x^{2} + y^{2}) d A = \int_{0}^{2 x} \int_{2}^{10} \sin (r^{2}) \cdot r d r d θ : p u t r^{2} = u \Rightarrow 2 r d r = d u \\ \Rightarrow r d r = \frac{d u}{2} \\ 2 \leq r \leq 8 \Rightarrow 4 \leq u \leq 64 \\ = [θ]_{0}^{2 x} \int_{4}^{64} \sin (u) \cdot \frac{d u}{2} \\ = 2 x \cdot \frac{1}{2} [- \cos (u)]_{4}^{64} \\ x [\cos (4) - \cos (64)] \end{matrix}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?