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Mathematics 4 Online
OpenStudy (anonymous):

Find the surface area of a sphere of radius r, using calculus.

myininaya (myininaya):

omg i know this

myininaya (myininaya):

give me just a sec

OpenStudy (anonymous):

that would be amazing if you could help! :)

myininaya (myininaya):

so the equation of a circle having center (0,0) is x^2+y^2+r^2

myininaya (myininaya):

oh surface area not volume one sec. let me rething my strategy

OpenStudy (anonymous):

ok. thanks

OpenStudy (anonymous):

Use the formula for the surface area of a general equation revolved around the x-axis. If you start with x^2 + y^2 = r^2, and your function is y, then f(x) = sqrt(r^2-x^2). The formula for the surface area of revolution is, in this case, \[SA = 2 \pi* \int\limits_{-r}^{r} \ \ f(x) * \sqrt{1+f(x)^2}\] Integrate and simplify.

OpenStudy (anonymous):

Sorry, inside the root of the integrand it should be f'(x)^2.

myininaya (myininaya):

what is the circumference of sphere

OpenStudy (anonymous):

Thanks QuantumModulus, that helped a lot!!

OpenStudy (anonymous):

Glad to help. :)

myininaya (myininaya):

have you done trig substition yet?

myininaya (myininaya):

I think you may have to do in one of the steps

myininaya (myininaya):

nvm it is a simple integration

myininaya (myininaya):

myininaya (myininaya):

ok but i think was suppose to get 4pi*r^2

myininaya (myininaya):

let me know what you get?

myininaya (myininaya):

ok it is 4*pi*r^2 because the semicirlce is being revolved about the x axis

myininaya (myininaya):

Yes! :)

myininaya (myininaya):

ignore the extra stuff on that attachment

myininaya (myininaya):

I have to go. You can check your work with the attachment just ignore the But part. goodnight

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