Question

find the area
a. y^2=-x; x=-2; x=-4
b. x^2 +y+4=0; y=-8 Take the elements of area parallel to the y-axis.
c. x^2 -y+1=0; x-y+1=0 Take the elements parallel to the x-axis.
d. x=4-y^2; x=4-4y

Answer 1

Find the area:

a) $y^{2} = -x, x = -2, x = -4$

b) $x^{2} + y + 4 = 0, y = -8$. Take the elements of area parallel to the y-axis.

$$S = 2 \int_{0}^{z} (-x^{2} - 4 + 8) dx = 2 \left(4x - \frac{x^{3}}{3}\right) \Big|_{-0}^{2} = \frac{32}{3}$$

c) $x^{2} - y + 1 = 0, x - y + 1 = 0$ . Take the elements parallel to the x-axis.

$x^{2} - y + 1 = x - y + 1$

$x = 0$ and $x = 1$

When $x = 0$ , then $y = 1$

When $x = 1$ , then $y = 2$

$S = \int_{1}^{2}(\sqrt{y - 1} - y + 1)dy = \left(\frac{2}{3}\sqrt{(y - 1)} - \frac{1}{2} y^2 + y\right)\bigg|_{1}^{2} = \frac{2}{3} - 2 + 2 + \frac{1}{2} - 1 = \frac{1}{6}$

d) $x = 4 - y^2$ , $x = 4 - 4y$

$$4 - y ^ {2} = 4 - 4 y$$ $$y ^ {2} - 4 y = 0$$ $$y = 0 \text{ and } y = 4$$ $$S = \int_ {0} ^ {4} (4 - y ^ {2} - 4 + 4 y) d y = \left(- \frac {1}{3} y ^ {3} + 2 y ^ {2}\right) \Big | _ {0} ^ {4} = - \frac {64}{3} + 32 = \frac {32}{3}$$