Find the area bounded by the functions y = 2x^{3} and y = 2x^{\frac{1}{3}} . 1.) Find points of intersection: 2x^{3} = 2x^{\frac{1}{3}} implies x^{3} = x^{\frac{1}{3}} implies x^{9} = x. Thus x^{9} − x = x(x^{8} − 1) = 0. Hence x = 0 and x^{8} − 1 = 0. x^{8} = 1 implies x = 1, −1 Hence the functions y = 2x^{3} and y = 2x^{\frac{1}{3}} intersect when x = −1, 0, 1 2.) Draw a rough graph:

3.) Find area: Method 1: Use vertical rectangles: ∫_{−1}^{0}[2x^{3} − 2x^{\frac{1}{3}} ]dx + ∫_{0}^{1}[2x^{\frac{1}{3}} − 2x^{3}]dx

Method 2: Use horizontal rectangles:

If y = 2x^{3}, then x = (\frac{y}{2} )^{\frac{1}{3}} . If y = 2x^{\frac{1}{3}} , then x = (\frac{y}{2} )^{3}. When x = −1, y = −2. When x = 0, y = 0. When x = 1, y = 2. ∫_{−2}^{0}[(\frac{y}{2} )^{3} − (\frac{y}{2} )^{\frac{1}{3}} ]dy + ∫_{0}^{2}[(\frac{y}{2} )^{\frac{1}{3}} − (\frac{y}{2} )^{3}]dy Find the volume of the object obtained by rotating the area bounded by the functions y = 2x^{3} and y = 2x^{\frac{1}{3}} about the x−axis (y = 0):

π ∫_{−1}^{0}[(2x^{\frac{1}{3}} )^{2} − (2x^{3})^{2}]dx + π ∫_{0}^{1}[(2x^{\frac{1}{3}} )^{2} − (2x^{3})^{2}]dx = ∫_{−1}^{1}[(2x^{\frac{1}{3}} )^{2} − (2x^{3})^{2}]dx

Find the volume of the object obtained by rotating the area bounded by the functions y = 2x^{3} and y = 2x^{\frac{1}{3}} . about y = 4:

π ∫_{−1}^{0}[(4 − 2x^{\frac{1}{3}} )^{2} − (4 − 2x^{3})^{2}]dx + π ∫_{0}^{1}[(4 − 2x^{3})^{2} − (4 − 2x^{\frac{1}{3}} )^{2}]dx

Find the volume of the object obtained by rotating the area bounded by the functions y = 2x^{3} and y = 2x^{\frac{1}{3}} . about y = −2:

π ∫_{−1}^{0}[(2x^{3} − (−2))^{2} − (2x^{\frac{1}{3}} − (−2))^{2}]dx + π ∫_{0}^{1}[(2x^{\frac{1}{3}} − (−2))^{2} − (2x^{3} − (−2))^{2}]dx

= π ∫_{−1}^{0}[(2x^{3} + 2)^{2} − (2x^{\frac{1}{3}} + 2)^{2}]dx + π ∫_{0}^{1}[(2x^{\frac{1}{3}} + 2)^{2} − (2x^{3} + 2)^{2}]dx Find the volume of the object obtained by rotating the area bounded by the functions y = 2x^{3} (x = (\frac{y}{2} )^{\frac{1}{3}} )and y = 2x^{\frac{1}{3}} (x = (\frac{y}{2} )^{3}) about the y−axis (x = 0):

π ∫_{−2}^{0}[((\frac{y}{2} )^{\frac{1}{3}} )^{2} − ((\frac{y}{2} )^{3})^{2}]dy + π ∫_{0}^{2}[((\frac{y}{2} )^{\frac{1}{3}} )^{2} − ((\frac{y}{2} )^{3})^{2}]dy = π ∫_{−2}^{2}[((\frac{y}{2} )^{\frac{1}{3}} )^{2} − ((\frac{y}{2} )^{3})^{2}]dy

Find the volume of the object obtained by rotating the area bounded by the functions y = 2x^{3} (x = (\frac{y}{2} )^{\frac{1}{3}} )and y = 2x^{\frac{1}{3}} (x = (\frac{y}{2} )^{3}) about x = 8:

π ∫_{−2}^{0}[(8 − (\frac{y}{2} )^{\frac{1}{3}} )^{2} − (8 − (\frac{y}{2} )^{3})^{2}]dy + π ∫_{0}^{2}[(8 − (\frac{y}{2} )^{3})^{2} − (8 − (\frac{y}{2} )^{\frac{1}{3}} )^{2}]dy

Find the volume of the object obtained by rotating the area bounded by the functions y = 2x^{3} (x = (\frac{y}{2} )^{\frac{1}{3}} )and y = 2x^{\frac{1}{3}} (x = (\frac{y}{2} )^{3}) about x = −1:

π ∫_{−2}^{0}[((\frac{y}{2} )^{3} − 1)^{2} − ((\frac{y}{2} )^{\frac{1}{3}} − 1)^{2}]dy + π ∫_{0}^{2}[((\frac{y}{2} )^{\frac{1}{3}} − 1)^{2} − ((\frac{y}{2} )^{3} − 1)^{2}]dy