Question

set up iterated integrals for both orders of integration. then evaluate the double integral using the easier order and explain why it's easier.

_D y dA, D is bounded by y = x - 2, x=y²   (the D next to the double integral should be under the integral. I don't know how to put it in the right spot.  

Answer 1

 

$\begin{align*}&{\color{blue}(1)}\\\\& \begin{aligned}&\! y = x - 2 && \Rightarrow && x = y + 2 \\\\& \! x = y^2 && \Rightarrow && y = \pm x^{1/2} \end{aligned} \end{align*}$


$\begin{align*}&{\color{blue}(2)}\\\\& \begin{aligned}&\! y + 2 = y^2 &\Rightarrow\\\\&\! y^2 - y - 2 = 0 &\Rightarrow\\\\&\! (y - 2)(y + 1) = 0 &\Rightarrow\\\\&\! y = -1 \text{, } 2 \end{aligned} \end{align*}$


$\begin{align*}&{\color{blue}(3)}\\\\& \begin{aligned}\! I &= \iint_D y \; \text{d}A \\\\&= \int_{-1}^2 \int_{y^2}^{y + 2} y \; \text{d}x \, \text{d}y \\\\&= \int_0^1 \int_{-x^{1/2}}^{x^{1/2}} y \; \text{d}y \, \text{d}x \;+\; \int_1^4 \int_{x - 2}^{x^{1/2}} y \; \text{d}y \, \text{d}x \end{aligned} \end{align*}$


$\begin{align*}&{\color{blue}(4)}\\\\& \text{If you integrate with respect to } x \text{ first, then you only} \\& \text{have to evaluate one iterated integral.} \\\\& \begin{aligned}\! I &= \int_{-1}^2 \int_{y^2}^{y + 2} y \; \text{d}x \, \text{d}y \\\\&= \int_{-1}^2 xy \; \bigg|_{y^2}^{y + 2} \; \text{d}y \\\\&= \int_{-1}^2 y(y + 2 - y^2) \; \text{d}y \\\\&= \int_{-1}^2 y^2 + 2y - y^3 \; \text{d}y \\\\&= \frac{1}{3}y^3 + y^2 - \frac{1}{4}y^4 \; \bigg|_{-1}^2 \\\\&= (\tfrac{8}{3} + 4 - 4) - (-\tfrac{1}{3} + 1 - \tfrac{1}{4}) \bigg . \\\\&= \frac{9}{4} \end{aligned} \end{align*}$