Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
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Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
calculus
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up vote
1
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Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
calculus
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
calculus
Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
calculus
calculus
asked Apr 26 '17 at 3:40
tortugaaaaaaaa
62
62
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2 Answers
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The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation
$$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
and ultimately to $b=colorred2(2-sqrt[3]2)$.
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You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
$$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
and you can proceeded here.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation
$$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
and ultimately to $b=colorred2(2-sqrt[3]2)$.
add a comment |Â
up vote
0
down vote
The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation
$$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
and ultimately to $b=colorred2(2-sqrt[3]2)$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation
$$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
and ultimately to $b=colorred2(2-sqrt[3]2)$.
The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation
$$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
and ultimately to $b=colorred2(2-sqrt[3]2)$.
answered Apr 26 '17 at 3:47
Jack D'Aurizioâ¦
274k32268638
274k32268638
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up vote
0
down vote
You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
$$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
and you can proceeded here.
add a comment |Â
up vote
0
down vote
You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
$$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
and you can proceeded here.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
$$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
and you can proceeded here.
You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
$$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
and you can proceeded here.
answered Apr 26 '17 at 3:51
Nosrati
22.1k51747
22.1k51747
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