# Math Help - Find the surface area generated by revolving the curve

1. ## 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

2. Originally Posted by macadamianchica
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]

3. Originally Posted by macadamianchica
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.

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

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

You may or may not want to include the circular side faces.

4. Originally Posted by macadamianchica
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

the funct y=sqrt25-x^ is a halfcircle. find the integral by replacing x=5sint t. then u have sqrt of 25-25sin^2 x=sqrt25(1-sin^2 x)=5cost now find the definite integral of y=5cost

5. This is not the "integration of a function problem".
It's a 3-dimensional "surface of revolution".