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3 years ago

PLEASE help me with this question!! I really need help!

Mathematics

2 answers:

pochemuha3 years ago

6 0

First, use the volume of sphere formula by plugging in the volume into the equation to get the value of r cubed. Then get r by determining the cube root of r cubed. Once you have the value of r, plus that into the surface area of sphere formula to get the surface area of the balloon. Answer: 324pi inches squared (use radian mode on calculator)

alexdok [17]3 years ago

5 0

Answer:

324π in²

Step-by-step explanation:

The surface area of a sphere = 4πr² ( r is the radius )

Calculate r using the volume (V) formula

V = $\frac{4}{3}$ πr³

Here V = 972π , thus

$\frac{4}{3}$ πr³ = 972π ( divide both sides by π )

$\frac{4}{3}$ r³ = 972 ( multiply both sides by 3 to clear the fraction )

4r³ = 2916 ( divide both sides by 4 )

r³ = 729 ( take the cube root of both sides )

r = $\sqrt[3]{729}$ = 9

Thus

surface area = 4π × 9² = 4π × 81 = 324π in²

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Bas_tet [7]

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3 years ago

Read 2 more answers

CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS

finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve $y=x^2$ , with $1\le x\le2$ , and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

$\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6$

But perhaps you'd like the surface integral treatment. Parameterize the surface by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$ , where the third component follows from

$z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2$

Take the normal vector to the surface to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k$

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

$\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}$

Then the area of the surface is

$\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$