Calculate the exact area of the region
(a) $4x = 4y - y^2, 4x - y = 0$
My solution:
$4x = 4y - y^2$
(1) $x = \frac{1}{4}y(4-y)$
4x = y
(2) $x = \frac{1}{4}y$
Intercepts:
$\frac{1}{4}y = \frac{1}{4}(4-y)$
$0 = \frac{1}{4}y(3-y)$
y = 0 and y = 3
Because the function is positive on [0,3]
$$=\frac{1}{4}\int_{0}^{3} (3y-y^2)dy$$
$$=\big(\frac{3}{8}y^2 - \frac{1}{12}y^3\big) \bigg|_{0}^{3} = \frac{9}{8}$$
(b) $y = ln(x), y = 1, y = -1, x = y^2-2$ \
(1) y = ln(x)
(2) y = 1
(3) y = -1
(4) $y = \sqrt{x+2}$
(5) $y = -\sqrt{x+2}$
I have no idea how to do b), and is my solution for a correct? Thx
Okay I was thinking for b, could I do this with respect to y by letting x = e^y, would this get rid of the y = 1 and -1? Then again, how am I going to get the intercepts for e^y and $y^2-2$
Maybe y = 1 and y = -1 can you used as the upper / lower limits?