Q. 3 - Calculus

Question

Evaluating a double integral as an iterated integral: Use Fubini’s theorem to evaluate the given double integrals. For each integral, show that you obtain the same result when you integrate using both possible orders of integration when R = {(x, y) | 0 ≤ x ≤ 2 and 1 ≤ y ≤ 4}. 

$\iint (x+2y)dA$

Step-by-Step Solution

Answer

$\iint (x+2y)dA=36 Square units$

Step 1: Fubini’s theorem

Let a< b and c<d be real numbers, let R be the rectangle defined by,

$R=\left\{(x,y)|a\le x\le b and c\le y\le d\right\}$ and let f(x) is continuous on R, Then

$\iint f(x,y)dA={\int }_a^b{\int }_c^{d}f(x,y)dxdy$

Step 2: Evaluate the integral

$$\begin{gathered} \iint (x+2y)dA={\int }_1^4{\int }_0^2(x+2y)dxdy \\ ={\int }_1^4\left[{\int }_0^2(x+2y)dx\right]dy \\ ={\int }_1^{4}dy{\left[\frac{x^2}{2}+2xy\right]}_0^2 \\ ={\int }_1^{4}dy\left[\frac{2^2}{2}+2\times 2\times y\right] \\ ={\int }_1^4(2+4y)dy \\ ={\left[2y+4\frac{y^2}{2}\right]}_1^4 \\ =\left(8+32\right)-\left(2+2\right) \\ =40-4 \\ =36 \end{gathered}$$