Find the volume of the solid obtained by rotating the region bounded by y=x^3, y=1, and the y-axis around the x-axis.
What I did was:
V=�ç(0,1)pi(1-x^3)^2 dx
v=pi(x^7/7-x^4/2+x)
and I evaluated it for one since 0 is just going to be 0.
and I got 9pi/14
The answer is wrong, can someone please explain me step by step I really don't know what's wrong.
How about taking the cylinder from 0 to 1 with radius 1 - the region below y = x^3
= π(1^2)(1) - π∫x^3 dx
= π - π[(1/4)x^4 ] from 0 to 1
= π - π(1/4 - 0) = (3/4)π
You interpret the washers wrong. If you have a disc of radius R with a hole in it of radius r, the area is
πR^2 - πr^2, not π(R-r)^2
forgot to square the radius ....
V = π - π∫(x^3)^2 dx
= π - π[(1/7)x^7 ] from 0 to 1
= π - π(1/7 - 0) = (6/7)π
To find the volume of the solid obtained by rotating the region bounded by y = x^3, y = 1, and the y-axis around the x-axis, you need to use the method of cylindrical shells.
The formula to find the volume of a solid obtained by rotating a region bounded by y = f(x), y = g(x), and the x-axis around the x-axis is given by:
V = ∫[a,b] 2πx(f(x) - g(x)) dx
In this case, the region is bounded by y = x^3, y = 1, and the y-axis, which gives us the limits of integration as [0, 1]. We will integrate with respect to x.
So, the volume equation becomes:
V = ∫[0,1] 2πx((x^3) - 1) dx
Now, let's evaluate this integral step-by-step:
V = 2π ∫[0,1] (x^4 - x) dx
Integrating term by term:
V = 2π [(1/5)x^5 - (1/2)x^2] |[0,1]
Now, let's substitute the limits of integration:
V = 2π [(1/5)(1^5) - (1/2)(1^2)] - [(1/5)(0^5) - (1/2)(0^2)]
Simplifying:
V = 2π [(1/5) - (1/2)] - [0 - 0]
V = 2π [-3/10]
V = -6π/10
Simplifying further:
V = -3π/5
So the correct answer for the volume of the solid obtained by rotating the region is -3π/5, which is approximately -1.88496.
It seems like there was a sign error in your calculations, resulting in the incorrect answer. Make sure to double-check your integrals and signs while evaluating them.