Find the surface area of a hemisphere that has a volume of 486π

title="cm^{3}" alt="cm^{3}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Xelga [282]2 years ago

6 0

$\textit{volume of a hemisphere}\\ V=\cfrac{4\pi r^3}{3}\cdot \cfrac{1}{2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=486\pi \end{cases}\implies 486\pi =\cfrac{4\pi r^3}{6} \implies 6(486\pi )=4\pi r^3 \\\\\\ \cfrac{6(486\pi )}{4\pi }=r^3\implies 729=r^3\implies \sqrt[3]{729}=r\implies \boxed{9=r} \\\\[-0.35em] ~\dotfill$

$\textit{surface area of a hemisphere}\\\\ SA=4\pi r^2\cdot \cfrac{1}{2}\implies \stackrel{\textit{we know that r = 9}}{SA=2\pi (9)^2}\implies SA=162\pi \implies SA\approx 508.94$

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$f(x) = \dfrac4{\sqrt{3-x}}$

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$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

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