Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves x = 0 , y = 1 , x = y^ 3 , about the line y = 1

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This is a great question, as it will involve a solid of revolution. The three main techniques for calculating solids of revolution are the shell method, the disk method, and the washer method. Our first step is to determine which one to use. Generally, when you’re integrating with respect to y, the shell method (which involved a series of concentric shells stacked inside of one another) is appropriate. Otherwise the best method is the disk or washer method. Since this problem involves rotating around a horizontal line (y=1) and there is no inner circle taken out of the solid (resulting in the shape of a washer), we can use the disk method.

This will involve first finding the intersection of the lines y=1 and y=x^(1/3). The intersection point is (1,1). We have x=0 as our lower bound. Next, since we are rotating around the line y=1, we can rewrite the integral as the radius of the circles that will be stacked together to create a solid by means of integration. The integrand will look like (1-x^(1/3))^2 and we will put a pi in front of the integral. The “squared” results from the formula for the area of a circle (pi*r^2). Recall that the function (1-x^(1/3)) defines the ever-changing radius of this solid. If you integrate (manually or with a software) using the lower bound of 0 and an upper bound of 1, you will yield a volume of pi/10 (or about .314).

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