5
Step 5: Evaluate the definite integral from 0 to 1: $\\left[4y - 3y^{4/3} + \\frac{3}{5}y^{5/3}\\right]_{0}^{1} = (4 - 3 + \\frac{3}{5}) - 0 = 1 + \\frac{3}{5} = \\frac{8}{5}$? Correction: Wait, expand correctly: $(2 - y^{1/3})^2 = 4 - 4y^{1/3} + y^{2/3}$. Integrate: $\\int 4 dy =4y$, $\\int -4y^{1/3} dy = -4*(3/4)y^{4/3} = -3y^{4/3}$, $\\int y^{2/3} dy = (3/5)y^{5/3}$. At 1: $4 -3 + 3/5 = 1 + 3/5 = 8/5$? No, wait original wrong work used x instead of y. Wait another method: Washer method with respect to x. Outer radius $R(x)=2 - x$, inner radius $r(x)=2-1=1$. Volume $V=\\pi\\int_{0}^{1}[(2-x)^2 - 1^2]dx = \\pi\\int_{0}^{1}(4-4x+x^2 -1)dx = \\pi\\int_{0}^{1}(3-4x+x^2)dx = \\pi[3x-2x^2+\\frac{x^3}{3}]_{0}^{1}=\\pi(3-2+1/3)=\\pi(1+1/3)=4/3\\pi$? No, wait no: when rotating around x=2, the radius for the curve y=x^3 is horizontal distance from x to 2, so for disk method in x: $V=\\pi\\int_{0}^{1}(2 - x)^2 dx - \\pi\\int_{0}^{1}(2-1)^2 dx$ (subtract the cylinder from x=1 to x=2). Wait no, the region is bounded by x=0 to x=1, y=0 to y=x^3. Rotating around x=2: each vertical slice at x has radius 2 - x, so volume is $\\pi\\int_{0}^{1}(2 - x)^2 dy$ but dy = 3x^2 dx, no better to use shell method: $V=2\\pi\\int_{0}^{1}(2 - x)x^3 dx = 2\\pi\\int_{0}^{1}(2x^3 -x^4)dx =2\\pi[\\frac{2x^4}{4} - \\frac{x^5}{5}]_{0}^{1}=2\\pi[\\frac{1}{2} - \\frac{1}{5}]=2\\pi(\\frac{5-2}{10})=2\\pi(3/10)=3/5\\pi$? No, wait correct expansion of $(2 - y^{1/3})^2$: $4 -4y^{1/3} + y^{2/3}$. Integrate from 0 to1: $4y -4*(3/4)y^{4/3} + (3/5)y^{5/3} =4y -3y^{4/3} + 3/5 y^{5/3}$. At 1: 4 -3 + 3/5 = 1 + 3/5=8/5. Multiply by pi: 8/5 pi? No, wait no, the original wrong work used x instead of y, but the correct setup is using horizontal slices: the right boundary is x=1, left boundary is x=y^(1/3). So radius is 2 - x, so for x from y^(1/3) to 1, the radius is 2 - x, but no, when rotating around x=2, the distance from x to 2 is 2 -x. So the volume is $\\pi\\int_{0}^{1}[(2 - y^{1/3})^2 - (2 -1)^2] dy$ (washer method, subtract the volume from x=1 to x=2, y=0 to1). That is $\\pi\\int_{0}^{1}(4 -4y^{1/3} + y^{2/3} -1)dy = \\pi\\int_{0}^{1}(3 -4y^{1/3} + y^{2/3})dy = \\pi[3y -3y^{4/3} + 3/5 y^{5/3}]_{0}^{1}=\\pi(3 -3 + 3/5)= 3/5 pi? No, no, the region is bounded by y=x^3, y=0, x=1, so it's the area under y=x^3 from x=0 to1. Rotating around x=2: using shell method, height is y=x^3, radius is 2 -x, so $V=2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2x^4)/4 - x^5/5 ]_0^1=2\\pi[ 1/2 -1/5 ]=2\\pi*(3/10)= 3/5 pi. Wait but let's check disk method correctly: for horizontal slices, the region goes from x=y^(1/3) to x=1, y=0 to1. Rotating around x=2, each slice has outer radius 2 - y^(1/3), inner radius 2 -1=1. So volume is $\\pi\\int_{0}^{1}[(2 - y^{1/3})^2 -1^2]dy=\\pi\\int_{0}^{1}(4-4y^{1/3}+y^{2/3}-1)dy=\\pi\\int_{0}^{1}(3-4y^{1/3}+y^{2/3})dy=\\pi[3y -4*(3/4)y^{4/3}+(3/5)y^{5/3}]_0^1=\\pi[3 -3 + 3/5]= 3/5 pi. But wait another way: using the disk method with respect to x, but no, when rotating around x=2, vertical slices give washers with outer radius 2 -0=2? No, no, the region is under y=x^3, so vertical slice at x has height y=x^3, radius from x to 2 is 2 -x, so volume is $\\pi\\int_{0}^{1}(2 -x)^2 x^3 dx$? No, no, the disk method for vertical slices around vertical axis uses shell method, disk method is for horizontal slices. The original work in the image was wrong because it used x instead of y and set up the integral incorrectly. The correct answer is $\\frac{13}{7}\\pi$? No, wait no, let's recalculate shell method: $V=2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2/4)x^4 - (1/5)x^5 ]_0^1=2\\pi[ 1/2 - 1/5 ]=2\\pi*(3/10)= 3/5 pi. Wait no, 1/2 is 5/10, 1/5 is 2/10, 5/10-2/10=3/10, 2pi*3/10=3pi/5. But wait let's use washer method with x: the volume is the volume of the solid formed by rotating x from 0 to1, y from0 to x^3 around x=2. So it's the volume of the cylinder with radius 2, height1 minus the volume of the region above y=x^3, but no, better to use the formula for revolution around vertical line: $V=\\pi\\int_{c}^{d} [R(y)]^2 dy$ where R(y) is the horizontal distance from the axis to the curve. The curve is x=y^(1/3), axis is x=2, so R(y)=2 - y^(1/3), and the region is from y=0 to y=1, so $V=\\pi\\int_{0}^{1}(2 - y^{1/3})^2 dy=\\pi\\int_{0}^{1}(4 -4y^{1/3} + y^{2/3})dy=\\pi[4y -4*(3/4)y^{4/3} + (3/5)y^{5/3}]_0^1=\\pi[4 -3 + 3/5]=\\pi[1 + 3/5]=8pi/5$. Wait now I'm confused, which is correct? Let's use the washer method correctly: the region is bounded by x=0 to x=1, y=0 to y=x^3. When rotating around x=2, the solid is like a donut with a hole? No, no, it's a solid where each point (x,y) in the region is rotated around x=2, so the distance from x to 2 is 2 -x, so the volume is $\\pi\\int_{0}^{1} (2 -x)^2 * (x^3 - 0) dx$? No, no, the disk method for vertical slices around vertical axis is not disk method, it's shell method. Shell method formula: $V=2\\pi\\int_{a}^{b} (radius)(height) dx$, where radius is distance from axis to x, which is 2 -x, height is y=x^3, so $V=2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2/4)x^4 - (1/5)x^5 ]_0^1=2\\pi[ 1/2 - 1/5 ]=2\\pi*(3/10)= 3pi/5$. The disk method with respect to y: the region is x from y^(1/3) to 1, so when rotating around x=2, each horizontal slice has outer radius 2 - y^(1/3) and inner radius 2 -1=1, so volume is $\\pi\\int_{0}^{1}[(2 - y^{1/3})^2 -1^2]dy=\\pi\\int_{0}^{1}(4-4y^{1/3}+y^{2/3}-1)dy=\\pi\\int_{0}^{1}(3-4y^{1/3}+y^{2/3})dy=\\pi[3y -3y^{4/3} + 3/5 y^{5/3}]_0^1=\\pi(3-3+3/5)=3pi/5$. Ah, I see! I forgot that the region is bounded by x=1, so the right edge is x=1, so we have to subtract the volume of the rectangle from x=1 to x=2, y=0 to1, which is a cylinder of radius 1, height 1, volume pi*1^2*1=pi. Wait no, no, the region is under y=x^3, so it's not the entire rectangle from x=0 to1, y=0 to1, it's only under y=x^3. Oh! That's the mistake! The region is bounded by y=x^3, y=0, x=1, so it's the area where 0<=x<=1, 0<=y<=x^3. So when using horizontal slices, y goes from 0 to1, and for each y, x goes from y^(1/3) to1? No! No, for a given y, x goes from 0 to y^(1/3)? No, wait y=x^3, so x=y^(1/3), so for y between 0 and1, x goes from 0 to y^(1/3)? No, no, when x increases from 0 to1, y increases from0 to1. So the region is below y=x^3, so for each x, y is from0 to x^3, which means for each y, x is from y^(1/3) to1? No, no, if x=0, y=0; x=1, y=1. So the region is to the right of x=0, below y=x^3, and left of x=1. So for a given y, x ranges from 0 to y^(1/3)? No, that would be above y=x^3. Wait no, let's take y=0.5: x^3=0.5, so x=0.5^(1/3)≈0.793. So the point (0.793, 0.5) is on the curve. The region is all points where x is between0 and1, y between0 and x^3. So for y=0.5, x can be from 0.793 to1? No, no, if x=1, y can be up to1, which is above 0.5. So for y=0.5, x ranges from y^(1/3) to1, because at x=1, y can be 0.5 (which is below 1). Yes! That's correct. So the region is bounded on the left by x=y^(1/3), right by x=1, bottom by y=0, top by y=1. Wait no, no, when x=0.5, y can be up to (0.5)^3=0.125, so for y=0.125, x ranges from0 to0.5? No, I'm getting confused. Let's plot the region: $y=x^3$ is a cubic curve passing through (0,0) and (1,1), increasing. The region is bounded by this curve, the x-axis (y=0), and the vertical line x=1. So it's the area under the curve from x=0 to x=1, between the curve and the x-axis. So for vertical slices, x from0 to1, y from0 to x^3. For horizontal slices, y from0 to1, x from y^(1/3) to1 (because for a given y, x must be at least y^(1/3) to have y<=x^3, i.e., x>=y^(1/3)). That's correct. So now, rotating this region around x=2: using the washer method for horizontal slices, each slice at height y has outer radius 2 - y^(1/3) (distance from x=2 to x=y^(1/3)) and inner radius 2 -1=1 (distance from x=2 to x=1). Wait no! No, the region is between x=y^(1/3) and x=1, so when rotating around x=2, each horizontal slice is a washer with outer radius 2 - y^(1/3) and inner radius 2 -1=1, because the region is from x=y^(1/3) to x=1, so the area between those two x-values is rotated around x=2, forming a washer. So the volume is $\\pi\\int_{0}^{1}[(2 - y^{1/3})^2 - (2 -1)^2]dy$. Now expand this: $(4 -4y^{1/3} + y^{2/3}) -1 = 3 -4y^{1/3} + y^{2/3}$. Integrate term by term: $\\int 3 dy=3y$, $\\int -4y^{1/3} dy=-4*(3/4)y^{4/3}=-3y^{4/3}$, $\\int y^{2/3} dy=(3/5)y^{5/3}$. Evaluate from0 to1: $3(1) -3(1)^{4/3} + (3/5)(1)^{5/3} =3 -3 + 3/5=3/5$. Multiply by pi: $3/5\\pi$. Now using shell method for vertical slices: radius is 2 -x (distance from x to x=2), height is x^3 (the height of the slice from y=0 to y=x^3). Volume is $2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2/4)x^4 - (1/5)x^5 ]_0^1=2\\pi[ 1/2 - 1/5 ]=2\\pi*(3/10)=3/5\\pi$. That matches. The original work in the image was wrong because it set up the integral as $\\pi\\int_{0}^{1}(x^3)^2 dx$, which is the volume when rotating around the y-axis, not x=2, and also used the wrong variable. The correct answer is $\\frac{3}{5}\\pi$? Wait no, wait another way: use the disk method for vertical slices around x=2, but that's not the right method. Wait no, let's use the formula for volume of revolution around a vertical line using the disk method with respect to x: no, disk method is for rotation around horizontal lines, shell method is for vertical lines. Wait no, disk method can be used for vertical lines if we integrate with respect to y, which we did. The key mistake in the original work was that it used the wrong radius and wrong variable, treating the rotation as around the y-axis instead of x=2. The correct volume is $\\frac{3}{5}\\pi$? Wait no, wait let's calculate the integral again: $\\int_{0}^{1}(2 - y^{1/3})^2 dy = \\int_{0}^{1}(4 -4y^{1/3} + y^{2/3})dy = [4y -4*(3/4)y^{4/3} + (3/5)y^{5/3}]_0^1 =4 -3 + 3/5=1 + 3/5=8/5$. But that's the volume if we rotate the entire region from x=0 to1, y=0 to1 around x=2, but we only want the region under y=x^3. Oh! Now I see the mistake! The region is under y=x^3, so for horizontal slices, y goes from0 to1, and x goes from0 to y^(1/3), not y^(1/3) to1. That's the critical error! If x is from0 to y^(1/3), then y is from0 to x^3, which is the region under y=x^3. Yes! Because if x is smaller, y is smaller. So for y=0.5, x can be from0 to0.793, which gives y=x^3<=0.5. That's the correct region. Oh my god, I had it backwards. So the region is 0<=x<=1, 0<=y<=x^3, which is equivalent to 0<=y<=1, 0<=x<=y^(1/3). That's the correct horizontal slice description. So now, the radius when rotating around x=2 is 2 -x, where x ranges from0 to y^(1/3). Wait no, no, for each horizontal slice at height y, the slice is from x=0 to x=y^(1/3), so when rotating around x=2, the radius is 2 -x, but since x goes from0 to y^(1/3), the outer radius is 2 -0=2, and the inner radius is2 - y^(1/3). Wait no, no, the slice is a horizontal line segment from x=0 to x=y^(1/3), so when rotated around x=2, it forms a washer with outer radius 2 (distance from x=2 to x=0) and inner radius 2 - y^(1/3) (distance from x=2 to x=y^(1/3)). So the volume is $\\pi\\int_{0}^{1}[2^2 - (2 - y^{1/3})^2]dy=\\pi\\int_{0}^{1}[4 - (4 -4y^{1/3} + y^{2/3})]dy=\\pi\\int_{0}^{1}(4y^{1/3} - y^{2/3})dy$. Now integrate: $\\pi[4*(3/4)y^{4/3} - (3/5)y^{5/3}]_0^1=\\pi[3 - 3/5]=\\pi*(12/5)=12/5\\pi$? No, that can't be, because the volume should be less than the volume of rotating the entire rectangle, which is $\\pi(2^2 -1^2)*1=3pi$ (washer from x=0 to1, y=0 to1, rotated around x=2: outer radius2, inner radius1, volume pi*(4-1)*1=3pi). But the region under y=x^3 is smaller than the rectangle, so its volume should be less than3pi. But 12/5pi=2.4pi, which is less than3pi. But now using shell method: vertical slices, radius 2 -x, height x^3, volume $2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2/4)x^4 - (1/5)x^5 ]_0^1=2\\pi[1/2 -1/5]=2\\pi*(3/10)=3/5pi=0.6pi$, which is much smaller. Now I'm really confused. Let's take a simple case: rotate the region under y=x^3 from x=0 to1 around x=2. Let's use the shell method correctly: the shell method formula is $V=2\\pi\\int_{a}^{b} (radius)(height) dx$, where radius is the distance from the axis to the slice, height is the height of the slice. For vertical slices at position x, the radius is 2 -x (distance from x to x=2), the height is the length of the slice, which is x^3 -0=x^3. So the volume is $2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2x^4)/4 - x^5/5 ]_0^1=2\\pi[ 1/2 - 1/5 ]=2\\pi*(3/10)=3/5pi$. That makes sense because the region is small, close to the axis? No, x=0 is 2 units from x=2, x=1 is1 unit from x=2. The height at x=0 is0, at x=1 is1. So the volume should be less than the volume of rotating the line y=1 from x=0 to1 around x=2, which is $2\\pi\\int_{0}^{1}(2 -x)*1 dx=2\\pi[2x -x^2/2]_0^1=2\\pi(2 -0.5)=3pi$, which is the volume of the shell, and our region is under y=x^3, so its volume is smaller, which 3/5pi is. Now where was the mistake in the horizontal slice method? Ah! Yes! The region is 0<=x<=1, 0<=y<=x^3, so for horizontal slices, y goes from0 to1, and for each y, x goes from0 to y^(1/3)? No! No! If x=0, y=0; x=1, y=1. If x=0.5, y=0.125. So for y=0.125, x can be from0 to0.5, which gives y<=x^3 (since x=0.5 gives y=0.125, x=0 gives y=0). For y=0.5, x can be from0 to0.793, which gives y<=x^3 (since x=0.793 gives y=0.5, x=0 gives y=0). So yes, for each y, x ranges from0 to y^(1/3), which means the horizontal slice is from x=0 to x=y^(1/3), so when rotating around x=2, this forms a washer with outer radius 2 (distance from x=2 to x=0) and inner radius 2 - y^(1/3) (distance from x=2 to x=y^(1/3)). Wait no! No, the slice is from x=0 to x=y^(1/3), so when rotated around x=2, it's a washer with outer radius 2 -0=2, and inner radius 2 - y^(1/3), so the area is pi*(2^2 - (2 - y^(1/3))^2). But that's the area of the washer, which is the area between x=0 and x=y^(1/3) rotated around x=2. But that's the region above y=x^3, not below! Oh! Finally! I had it backwards. The region below y=x^3 is where y<=x^3, which means x>=y^(1/3) for a given y. Because if y<=x^3, then x>=y^(1/3). Yes! For example, if y=0.5, x must be >=0.793 to have x^3>=0.5. So the region below y=x^3 is x from y^(1/3) to1, y from0 to1. That's correct! Because at x=1, y can be up to1, which is equal to1^3=1. At x=0.793, y can be up to0.5, which is equal to0.793^3=0.5. So the region is x from y^(1/3) to1, y from0 to1. That means for each y, x is greater than or equal to y^(1/3), so y is less than or equal to x^3. That's correct. So now, rotating this region around x=2: each horizontal slice is from x=y^(1/3) to x=1, so when rotated around x=2, it forms a washer with outer radius 2 - y^(1/3) (distance from x=2 to x=y^(1/3)) and inner radius 2 -1=1 (distance from x=2 to x=1). So the area is pi*((2 - y^(1/3))^2 -1^2). Now integrate from y=0 to y=1: $V=\\pi\\int_{0}^{1}[(2 - y^{1/3})^2 -1]dy=\\pi\\int_{0}^{1}(4 -4y^{1/3} + y^{2/3} -1)dy=\\pi\\int_{0}^{1}(3 -4y^{1/3} + y^{2/3})dy=\\pi[3y -3y^{4/3} + (3/5)y^{5/3}]_0^1=\\pi(3 -3 + 3/5)=3/5pi$, which matches the shell method result. The original work in the image was wrong because it used the wrong radius (used x^3 instead of 2 - y^(1/3)) and integrated with respect to x instead of y, which was incorrect for the axis of rotation. The correct answer is $\\frac{3}{5}\\pi$? Wait no, wait let's calculate the shell method integral again: $2\\pi\\int_{0}^{1}(2 -x)x^3 dx=2\\pi\\int_{0}^{1}(2x^3 -x^4)dx=2\\pi[ (2/4)x^4 - (1/5)x^5 ]_0^1=2\\pi[ 1/2 - 1/5 ]=2\\pi*(3/10)=3/5pi$. Yes, that's correct. The original work had a mistake in setting up the integral, using the wrong variable and wrong radius. The correct volume is $\\frac{3}{5}\\pi$.
