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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?

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    (a) is correct. for (b) take it with respect to y -as you mentioned-, give us your answer.2017-02-05
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    (1) x = y^2-2 (2) x = e^y2017-02-05
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    Sorry I was writing I got bugged out. Anyways what I meant to say was, how would I calculate their intercepts mathematically? As you can see for a I just did $x_r - x_l$ and found the zero of this function to be the intercepts. This won't work for this function because If I were to let $x_r$ to be x=e^y, then it'd be difficult to find the zeros with x= y^2 - 2. I have no idea how to do this question. Suggestion, maybe? I'm sure there are many ways to do this question but I'm only known of one way at the moments and thats only with polynomials.2017-02-05
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    A question, does y = 1, and y = -1, turn into x = 1 and x = -1?2017-02-05

1 Answers 1

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Hint: $$\int_{-1}^1\Big(e^y-(y^2-2)\Big)dy=\frac{e^2-1}{e}+\dfrac{10}{3}$$

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    Massive hint. Thx2017-02-05
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    I wait your answer.2017-02-05
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    Once computated I would get $$=\frac{e^2-1}{e} + \frac{10}{3}$$2017-02-05
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    Ok. Good job...2017-02-05
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    Thank you. Need more practice still, but this helps :)2017-02-05
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    Your welcome...2017-02-05