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Home / Calculus I / Applications of Integrals / Volumes of Solids of Revolution/Method of Cylinders
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### Section 6-4 : Volume With Cylinders

For each of the following problems use the method of cylinders to determine the volume of the solid obtained by rotating the region bounded by the given curves about the given axis.

1. Rotate the region bounded by $$x = {\left( {y - 2} \right)^2}$$, the $$x$$-axis and the $$y$$-axis about the $$x$$-axis. Solution
2. Rotate the region bounded by $$\displaystyle y = \frac{1}{x}$$, $$\displaystyle x = \frac{1}{2}$$, $$x = 4$$and the $$x$$-axis about the $$y$$-axis. Solution
3. Rotate the region bounded by $$y = 4x$$ and $$y = {x^3}$$ about the $$y$$-axis. For this problem assume that $$x \ge 0$$. Solution
4. Rotate the region bounded by $$y = 4x$$ and $$y = {x^3}$$ about the $$x$$-axis. For this problem assume that $$x \ge 0$$. Solution
5. Rotate the region bounded by $$y = 2x + 1$$, $$y = 3$$ and $$x = 4$$ about the line $$y = 10$$. Solution
6. Rotate the region bounded by $$x = {y^2} - 4$$ and $$x = 6 - 3y$$ about the line $$y = - 8$$. Solution
7. Rotate the region bounded by $$y = {x^2} - 6x + 9$$ and $$y = - {x^2} + 6x - 1$$ about the line $$x = 8$$. Solution
8. Rotate the region bounded by $$\displaystyle y = \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}$$, $$\displaystyle y = 5 - \frac{1}{4}x$$, $$x = - 1$$ and $$x = 6$$ about the line $$x = - 2$$. Solution