# How do I find the volume of the solid generated by revolving the region bounded by y=x^2, y=0, and x=2 about the x-axis? The y-axis?

May 19, 2018

1)$V o l u m e = \pi {\int}_{0}^{2} {x}^{4} \cdot \mathrm{dx} = \left(\frac{32}{5}\right) \pi {\left(u n i t e\right)}^{3}$

2)$V o l u m e = \pi {\int}_{0}^{4} \left[{\left({2}^{2}\right)}_{2} - {\left({\sqrt{y}}^{2}\right)}_{1}\right] \cdot \mathrm{dy} = \pi {\int}_{0}^{4} \left[4 - y\right] \cdot \mathrm{dy} = 8 \pi {\left(u n i t e\right)}^{3}$

#### Explanation:

the rose region is revolving about the x-axis and y-axis 1)when the shaded region revolving a bout x-axis

$V o l u m e = \pi {\int}_{a}^{b} {y}^{2} \cdot \mathrm{dx}$

$V o l u m e = \pi {\int}_{0}^{2} {y}^{2} \cdot \mathrm{dx} = \pi {\int}_{0}^{2} {x}^{4} \cdot \mathrm{dx} = \pi {\left[\frac{1}{5} \cdot {x}^{5}\right]}_{0}^{2}$

$= \pi \left[\left(\frac{32}{5}\right) - 0\right] = \left(\frac{32}{5}\right) \pi {\left(u n i t e\right)}^{3}$

$V o l u m e = \pi {\int}_{d}^{c} \left[{\left({x}^{2}\right)}_{2} - {\left({x}^{2}\right)}_{1}\right] \cdot \mathrm{dy}$
$V o l u m e = \pi {\int}_{0}^{4} \left[{\left({2}^{2}\right)}_{2} - {\left({\sqrt{y}}^{2}\right)}_{1}\right] \cdot \mathrm{dy}$
$= \pi {\int}_{0}^{4} \left[4 - y\right] \cdot \mathrm{dy} = \pi {\left[4 y - \frac{1}{2} {y}^{2}\right]}_{0}^{4}$
$= \pi \left[16 - 8\right] = 8 \pi {\left(u n i t e\right)}^{3}$