How do you solve for R in $a= 2piR^2 + 2piRh$ ?

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Answer 1 · 2016-07-12T17:21:13

$R= -h/2 +- sqrt(h^2/4 + a/(2pi)$

Notice we have a quadratic in R here:

$2piR^2 + 2pihR - a =0$

$R = (-2pih +-sqrt((2pih)^2-4(2pi)(-a)))/(4pi)$

$R = (-2pih +-sqrt(4pi^2h^2+8pia))/(4pi)$

$R = -h/2 +- (sqrt(4pi^2h^2+8pia))/(4pi)$

In the second fraction, we can rewrite the denominator from $4pi$ to $sqrt(16pi^2)$

Our solution becomes:

$R = -h/2 +-sqrt((4pi^2h^2+8pia)/(16pi^2)) = -h/2 +- sqrt(h^2/4 + a/(2pi)$

Answer 2 · 2016-07-12T17:30:20

$R=(-2pih+-sqrt((2pih)^2+8pia))/(4pi)$

$a=2piR^2+2piRh$ is a quadratic equation in $R$ and can be written as $2piR^2+2piRh-a=0$ .

We can now solve for $R$ using quadratic formula $(-b+-sqrt(b^2-4ac))/(2a)$ and hence $R=(-2pih+-sqrt((2pih)^2+8pia))/(4pi)$

How do you solve for R in a= 2piR^2 + 2piRha= 2piR^2 + 2piRh?