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calc II
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?Find the surface area when Y = (a^2 - x^2)^(1/2) is revolved around the x-axis?
?Geometrically, what have you found?
Please help. Thank you.
?Geometrically, what have you found?
Please help. Thank you.
Char. Limit said:Could you show us what you have tried? It might give us an idea on where you need help.
Also, there is a template for homework posts. For future reference, please use it. It should appear by default whenever you start a new topic in the homework section.
calc II said:?Find the surface area when Y = (a^2 - x^2)^(1/2) is revolved around the x-axis?
?Geometrically, what have you found?
Please help. Thank you.
The derivative of Y = (a^2 - x^2)^(1/2) is given by (a^2 - x^2)^(-1/2) * (-2x).
The indefinite integral of Y = (a^2 - x^2)^(1/2) is given by (1/2) * (a^2 - x^2)^(3/2) + C.
The domain of Y = (a^2 - x^2)^(1/2) is all real numbers between -a and a, and the range is all real numbers between 0 and a.
Yes, the graph of Y = (a^2 - x^2)^(1/2) can be shifted or translated by adding or subtracting a constant value to the expression. This will shift the graph horizontally.
To find the area under a curve represented by Y = (a^2 - x^2)^(1/2), you can use the definite integral formula with appropriate limits of integration. This will give you the area between the curve and the x-axis.