Find out the answer to "Calculate the double integral. ∫∫2(1+x2)1+y2dA, where R={(x,y)∣0≤x≤2,0≤y≤1}"

e1s2kat26

Answered question

2021-10-14

Calculate the double integral.

\int \int \frac{2 (1 + x^{2})}{1 + y^{2}} d A,

where R = {(x, y) ∣ 0 \leq x \leq 2, 0 \leq y \leq 1}

Answer & Explanation

2abehn

Skilled2021-10-15Added 88 answers

Step 1
$It is given that R = {(x, y) | 0 \leq x \leq 2, 0 \leq y \leq 1} . We have to find$
$\int \int_{R} \frac{2 (1 + x^{2})}{1 + y^{2}} d A$
Notice that
$\int \int_{R} \frac{2 (1 + x^{2})}{1 + y^{2}} d A = \int_{{y = 0}}^{1} \int_{{x = 0}}^{1} \frac{2 (1 + x^{2})}{1 + y^{2}} d x d y$
$= \int_{0}^{1} [\frac{2}{1 + y^{2}} \cdot \int_{0}^{2} 1 + x^{2} d x] d y = \int_{0}^{1} [\frac{2}{1 + y^{2}} (2 + \frac{8}{3})] d y$
$= \int_{0}^{1} (\frac{28}{3 (1 + y^{2})}) d y$
$= \frac{28}{3} \cdot \int_{0}^{1} \frac{1}{1 + y^{2}} d y$
$= \frac{28}{3} {[\arctan (y)]}_{0}^{1} [∵ \int \frac{1}{1 + y^{2}} d y = \arctan (y)]$
$y = \frac{28}{3} \cdot \frac{π}{4}$
$= \frac{7 π}{3}$
Result
$\int_{0}^{1} \int_{0}^{2} \frac{2 (1 + x^{2})}{1 + y^{2}} d x d y = \frac{7 π}{3}$

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