Volumes Question (1 Viewer)

ds-nathan · Jul 6, 2009

Hello,

I've got this simple question from Patel that I cannot get the correct answer for. Maybe you guys can help. The question says: The area common to the curves y = x^2 and y^2 = x is revolved about the y-axis. Find the volume generated by using:
(a) the disc method
(b) the cylindrical shells.

For the disc method - my disc was pi (y - y^2) dy and for the cylindrical - my cylinder was 2 pi x (x^2) dx.

Thanks guys.

gurmies · Jul 6, 2009

For disc method:

$\Delta V=\pi (R^{2}-r^{2})\Delta y \,\, , where \,\, R=\sqrt{y} \,\, and \,\, r=y^{2} \\\\ V = \pi\lim_{\Delta y\to 0}\sum_{y=0}^{1}(y-y^{4})\Delta y \\\\ = \pi\int_{0}^{1}(y-y^{4})dy \\\\ = \pi\left [ \frac{y^{2}}{2}-\frac{y^{5}}{5} \right ]^{1}_{0} \\\\ = \frac{3\pi}{10}u^{3}$

For cylindrical shell method:

$\Delta V= 2\pi (x)(\sqrt x-x^{2})\Delta x \\\\ V = 2\pi\lim_{\Delta x\to 0}\sum_{x=0}^{1}(x^{\frac{3}{2}}-x^{3})\Delta x \\\\ = 2\pi\int_{0}^{1}(x^{\frac{3}{2}}-x^{3})dx \\\\ = 2\pi\left [ \frac{2x^{\frac{5}{2}}}{5}-\frac{x^{4}}{4} \right ]^{1}_{0} \\\\ = \frac{3\pi}{10}u^{3}$

Volumes Question (1 Viewer)

ds-nathan

New Member

gurmies

Drover

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