1. ## Double Integration

1. $\int_{D} \int_{D} y^3 dA
$
where D: (0,2), (1,1), (3,2).

If you graph D you will see that its a triangle with base y = 2 and sides y = x and y = 1/2x + 1/2.

Our teacher doesen't want us splitting our integrals up into so you have to do this from x = y and x = 2y-1 so you get

$\int_{1}^{2} \int_{y}^{2y-1} y^3 dx ~ dy
$

=
$\int_{1}^{2} xy^3$ $\bigg{|}_{x = y}^{x= 2y-1}~ dy$

= $\int_{1}^{2} (2y-1)y - y^4 dy$

=
$\int_{1}^{2} 2y^2-y - y^4 dy$

=
${\frac{2y^3}{3} - \frac {y^2}{2} - \frac{y^5}{5}}$ $\bigg{|}_{1}^{2}$

= $\frac{2}{3} * 8 - \frac {1}{2} * 4 - \frac{1}{5} * 32 - \frac{2}{3} + \frac{1}{2} + \frac{1}{5}$

= $\frac{14}{3} - \frac {3}{2} - \frac{31}{5}$

= $\frac{91}{30}$ Where did I go wrong? (answer should be $\frac{147}{20}$)

2. $\int_{D} \int_{D} 2x-y dA
$
where D: x^2 + y^2 = 4.

If you graph D you will see that its a circle with radius 2

= $4* \int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} 2x-y dy ~ dx
$

= $4 * \int_{0}^{2} 2xy-\frac{1}{2}y^2$ $\bigg{|}_{y = 0}^{y= \sqrt{4-x2}}~ dx$

= $4 * \int_{0}^{2} (2x\sqrt{4-x^3} - \frac{1}{2}(4-x^2) dx$

=
$4*\int_{0}^{2}$ $2x\sqrt{4-x^2} dx - 2*\int_{0}^{2}4-x^2 dx$

letting u = 4- x^2 and du = -2x dx you get

=
$4*\int_{0}^{2}$ $-\sqrt{u} du - 2*\int_{0}^{2}4-x^2 dx$

= $4 *(-\frac{2}{3} * {u}^{\frac{3}{2}}) \bigg{|}_{u = 0}^{u= 4} - 2* (4x -\frac{1}{3}x^3)\bigg{|}_{x = 0}^{x= 2}$

= $4 *(-\frac{2}{3} * {4}^{\frac{3}{2}}) - 2* (8-\frac{1}{3} * 8)$

= $4 *(-\frac{2}{3} * 8) - 2* (\frac{16}{3})$

= $4 *(-\frac{16}{3}) - \frac{32}{3}$

= -32 Where Did I go wrong? (answer should be 0)

2. the triangle is actually bounded by the lines, y = 2, y = -x + 2, and y = (1/2)x + 1/2

so your integral is: $\int_1^2 \int_{{\color{red}2 - y}}^{2y - 1}y^3~dx \, dy$