Follow the steps to compute the volume of the solid obtained by rotating the region bounded by y=x^(2)quad" and "quad y=6x about the line x=0 using the method of disks or washers. a. Using the method of disks or washers, set up the integral. {:[V=int_(a)^(b)◻],[" with "a=◻" and "b=]:}

Question

Follow the steps to compute the volume of the solid obtained by rotating the region bounded by  y=x^(2)quad" and "quad y=6x about the line  x=0 using the method of disks or washers. a. Using the method of disks or washers, set up the integral.  {:[V=int_(a)^(b)◻],[" with "a=◻" and "b=]:}

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Find Intersection Points: First, we need to find the points of intersection between the two curves y = x^2 and y = 6x to determine the limits of integration a and b. To do this, we set the two equations equal to each other and solve for x: x^2 = 6x x^2 - 6x = 0 x(x - 6) = 0 This gives us two solutions: x = 0 and x = 6. These are the limits of integration for our volume integral.Set Up Integral: Now, we will set up the integral using the method of disks or washers. Since we are rotating around the x-axis (x=0), the volume V of the solid is given by the integral of the area of the disks from x = a to x = b. The area of a disk is π times the square of the radius of the disk. The radius of each disk is the distance from the x-axis to the curve, which is the function value y.  For the region between y = x^2 and y = 6x, the outer radius R(x) is given by y = 6x, and the inner radius r(x) is given by y = x^2. The volume of the solid is the difference between the volume formed by rotating the outer curve and the volume formed by rotating the inner curve.  The volume integral is: V = π∫(R(x)^2 - r(x)^2) dx from x = a to x = b V = π∫(6x)^2 - (x^2)^2 dx from x = 0 to x = 6Simplify Integral: Now we simplify the integral: V = π∫(36x^2 - x^4) dx from x = 0 to x = 6Integrate Function: Next, we integrate the function with respect to x: V = π[∫36x^2 dx - ∫x^4 dx] from x = 0 to x = 6 V = π[(12x^3/3) - (x^5/5)] from x = 0 to x = 6Evaluate Integral: Now we evaluate the integral from x = 0 to x = 6: V = π[(12(6)^3/3) - (6^5/5)] - π[(12(0)^3/3) - (0^5/5)] V = π[(12*216/3) - (7776/5)] V = π[(864) - (1555.2)]Recalculate Volume: We calculate the numerical value: V = π[(864) - (1555.2)] V = π[-691.2] Since volume cannot be negative, there must be a mistake in the calculation. Let's go back and check the integration and evaluation.

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