# Find the surface area generated by revolving the curve

• Feb 16th 2010, 11:08 AM
Find the surface area generated by revolving the curve
Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you
• Feb 16th 2010, 12:13 PM
blertta
Quote:

Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

find it using the definite integral of y in [-3;4]
• Feb 16th 2010, 12:18 PM
Quote:

Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

As the curve revolves around the x-axis, each point on graph of the function traces out a circle, with f(x) as radius.

To calculate the surface area, the circle circumferences are integrated.

$\displaystyle R=f(x),\ -3\le{x}\le{4}$

Hence, the curved surface area is.... $\displaystyle 2\pi\int_{-3}^4f(x)dx\ =\ 2\pi\int_{-3}^4\sqrt{25-x^2}dx$

You may or may not want to include the circular side faces.
• Feb 17th 2010, 01:34 AM
blertta
Quote: