Find the volume of the solid formed

Question

Find the volume of the solid formed by rotating the region bounded by the given curves about the indicated axis.

y = x^3/4, x = 0, y = 1

 (in the first quadrant); about the y-axis

Answer 1

V = 2π∫_a^b r(x)h(x)dx <-- x=r(x)=radius, (1-x^3/4)=h(x)=height

V = 2π∫₀¹ x(1-x^3/4)dx

V = 2π∫₀¹ (x-x^7/4)dx

V = 2π * ((1/2)x² - (4/11)x^11/4) [0,1]

V = 2π[(1/2)(1)² - (4/11)(1)^11/4]

V = 2π[1/2 - 4/11]

V = π - 8π/11

V = 11π/11 - 8π/11

V = 3π/11 cubic units

Hope this helped!

Answer 2

We have two choices of method: washers or cylindrical shells.

Washers:

Because our representative strip of the region is horizontal, we need to integrate with respect to y (the strip is dy in height, ie infinitesimally tall. Because the region is entirely adjacent to our axis of rotation (the y-axis), this will be disks rather than washers.

We need to solve for x in terms of y, so since y = x^3/4 then x = y^4/3. That is our radius.

V = π ∫₀¹ (y^4/3)² dy = π ∫₀¹ y^8/3 dy = π [ 3/11 y^11/3 ]₀¹ = 3π / 11

Cylindrical shells:

Our representative strip is vertical, of infinitesimal width dx. Because it is parallel to our axis of rotation, we integrate an expression representing the lateral surface area of a cylinder, ie 2πrh. x = r (dist. from axis of rotation) and h = 1 - x^3/4.

V = 2π ∫₀¹ x(1 - x^3/4) dx = 2π ∫₀¹ x - x^7/4 dx = 2π [ 1/2x² - 4/11x^11/4]₀¹ = 3π / 11