how do you find the volume enclosed by x=3y and x=y^3 rotated about the y-axis? when y=>0
ooops, typing error
last line should have been
= 36pi/7 √3
or 27.97 which matches the answer Daman gave
These curves intersect at (0,0) and at (3 sqrt3 , sqrt 3)
in the area enclosed, x = 3y is bigger(to the right of) x = y^3
So we have a horizontal ring of height dy , inner radius y^3, and outer radius 3y
We can find the volume of the outer cone formed by x = 3y from y = 0 to y = sqrt 3 and then subtract the volume formed by x = y^3
First the x = 3y
Area of slice of height dy = pi x^2 dy
= pi *9y^2 dy
so
= 9 pi integral y^2 dy from 0 to sqrt 3
= 9 pi y^3/3 = 3 pi (sqrt 3)^3 = 9 pi sqrt 3 is about 49
Then subtract the x = y^3 contents
Area of slice dy = pi x^2 dy = pi y^6 dy
so
pi integral y^6 dy = pi y^7/7
= (pi/7)(sqrt3)^7 is about 21
so about 49 - 21 = 28
you should have problem calculating that the two graphs intersect at (0,0) and (3√3, √3)
recall that for the volume of a region rotated about the y-axis we would have
volume = pi(integral) x^2 dy from bottom y to top y of the region
so V = pi(integral)[9y^2 - y^6)dy from 0 to √3
= pi[3y^3 - 1/7 y^7] from 0 to √3
= pi(9√3 - 27/7 √3)
= 27pi/7 √3
Check my arithmetic, I am just scribbling on a piece of paper.
To find the volume enclosed by the curve x = 3y and x = y^3 when rotated about the y-axis, we can use the method of cylindrical shells. This involves integrating the volume of infinitely thin cylindrical shells formed by rotating a vertical strip along the y-axis.
First, let's sketch the curves x = 3y and x = y^3. The graphs of these curves intersect at (0, 0) and (1, 1).
To set up the integral, we need to determine the limits of integration. The volume is bounded by y = 0 (the x-axis) and the curve intersection point at (1, 1).
Next, we need to express the volume element (dV) in terms of y. For this, we consider a small vertical strip with height dy, located at a distance y from the y-axis. The circumference of the cylindrical shell is given by 2πr, where r is the x-coordinate of the curve at that particular y-value.
Since the curve equation is given as x = 3y, the radius (r) of the cylindrical shell is equal to 3y.
Now, we can express the volume of the cylindrical shell as dV = 2πy(3y)dy = 6πy^2dy.
To obtain the total enclosed volume, we integrate the volume element from y = 0 to y = 1:
V = ∫[0 to 1] 6πy^2dy
Evaluating this integral will give us the volume enclosed by the curves when rotated about the y-axis.
Note: If you're not familiar with integral calculus or evaluating integrals, it would be helpful to use software, graphing calculators, or online tools to perform the calculation.