Find the volume of the solid obtained by rotating the region bounded by y= x^2, y=0, and x=3 about the x-axis.

or, using shells, you can do

V = ∫[0,9] 2πrh dy
where r = y and h = 3-x
= 2π∫[0,9] y(3-√y) dy
= 2π(3/2 y^2 - 2/5 y^(5/2)) [0,9]
= 2π(3/2 * 81 - 2/5 * 243)
= 2π(243/10)
= 243π/5

To find the volume of the solid obtained by rotating the region bounded by y = x^2, y = 0, and x = 3 about the x-axis, you can use the method of cylindrical shells.

First, let's find the limits of integration for x. The region is bounded by x = 3, so the limits of integration for x will be from 0 to 3.

Next, let's consider a small strip or "cylindrical shell" with thickness Δx at a particular x-value. The radius of this cylindrical shell is x, and the height of the cylindrical shell is given by the difference between the y-values of the two curves at that x-value. Thus, the height of the cylindrical shell is y = x^2 - 0 = x^2.

The volume of a cylindrical shell is given by V = 2πrhΔx, where r is the radius and h is the height.

Substituting in the values, we have V = 2π(x)(x^2)(Δx) = 2πx^3Δx.

To find the volume of the entire solid, we need to integrate this expression with respect to x over the given limits of integration:

V = ∫[0,3] 2πx^3 dx

Evaluating this integral, we get:

V = 2π ∫[0,3] x^3 dx

Integrating x^3, we get:

V = 2π * [x^4/4] evaluated from 0 to 3

V = 2π * [(3^4/4) - (0^4/4)]

V = 2π * [(81/4) - 0]

V = 2π * (81/4)

Finally, simplifying, we have:

V = 81π/2

Therefore, the volume of the solid obtained by rotating the region bounded by y = x^2, y = 0, and x = 3 about the x-axis is 81π/2 cubic units.

To find the volume of the solid obtained by rotating a region about the x-axis, we can use the method of cylindrical shells.

First, we need to determine the limits of integration, which are the x-values of the region. From the given information, the region is bounded by y = x^2, y = 0, and x = 3. The limits of integration are therefore x = 0 to x = 3.

Next, we need to express the equations of the curves in terms of x. Since y = x^2 is already in terms of x, we can use it as it is.

Now, let's consider a small vertical strip of thickness Δx located at an arbitrary x-value within the interval [0, 3]. This strip will generate a cylindrical shell when rotated about the x-axis.

The height of this cylindrical shell is given by the difference in y-values between the curves y = x^2 and y = 0, which is simply x^2 - 0 = x^2.

The circumference of the cylindrical shell is given by the formula for the circumference of a circle, which is 2πr. In this case, the radius of the shell is the x-value of the strip.

The volume of each cylindrical shell can be calculated as the product of the height, circumference, and thickness Δx. So, the volume of the shell is (x^2)(2πx)(Δx) = 2πx^3(Δx).

To find the total volume of the solid, we need to integrate the expression for the volume of the cylindrical shells over the interval [0, 3].

V = ∫[0, 3] 2πx^3 dx

Integrating the function, we get:

V = 2π ∫[0, 3] x^3 dx

Evaluating this integral, we have:

V = 2π * [x^4/4] evaluated from 0 to 3

V = 2π * (3^4/4 - 0^4/4)

V = 2π * 81/4

V = 81π/2

Therefore, the volume of the solid obtained by rotating the region bounded by y = x^2, y = 0, and x = 3 about the x-axis is 81π/2 cubic units.

V = π∫(y^2) dx from a to b

= π∫x^4 dx from 0 to 3
= π[x^5/5] from 0 to 3
= π(243/5 - 0) = 243π/5 cubic units