how to calculate y = 9 - x ^ 2 interval [0, 2 ] I want to know solid volume by rotating 360 degrees around the x axis
+45 You should ignore what that person said because he/she said do not ask your Marist
I THINK it is Integral (0-2pi) Integral(0-2) 9-x^2 I think! 30 2/3 pi Yikes....that is a wild guess. Let me know when soemeone is able to ACTUALLY answer it .....
+128707 We are using the ring ["washer"] method........see the explanation, here....
http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx
[Example 1 is very similar to this one]
Compute the definite integral: integral_0^2 pi (9-x^2)^2 dx Factor out constants: = pi integral_0^2 (9-x^2)^2 dx Expanding the integrand (9-x^2)^2 gives x^4-18 x^2+81: = pi integral_0^2 (x^4-18 x^2+81) dx Integrate the sum term by term and factor out constants: = pi integral_0^2 x^4 dx+-18 pi integral_0^2 x^2 dx+81 pi integral_0^2 1 dx Apply the fundamental theorem of calculus. The antiderivative of x^4 is x^5/5: = (pi x^5)/5|_0^2+-18 pi integral_0^2 x^2 dx+81 pi integral_0^2 1 dx Evaluate the antiderivative at the limits and subtract. (pi x^5)/5|_0^2 = (pi 2^5)/5-(pi 0^5)/5 = (32 pi)/5: = (32 pi)/5+-18 pi integral_0^2 x^2 dx+81 pi integral_0^2 1 dx Apply the fundamental theorem of calculus. The antiderivative of x^2 is x^3/3: = (32 pi)/5+(-6 pi x^3)|_0^2+81 pi integral_0^2 1 dx Evaluate the antiderivative at the limits and subtract. (-6 pi x^3)|_0^2 = (-6 pi 2^3)-(-6 pi 0^3) = -48 pi: = -(208 pi)/5+81 pi integral_0^2 1 dx Apply the fundamental theorem of calculus. The antiderivative of 1 is x: = -(208 pi)/5+81 pi x|_0^2 Evaluate the antiderivative at the limits and subtract. 81 pi x|_0^2 = 81 pi 2-81 pi 0 = 162 pi: Answer: | | = (602 pi)/5 =~378.25
