View source for Calculus/Volume of solids of revolution/Solutions

{{question-answer|question=1. Calculate the volume of the cone with radius <math>r</math> and height <math>h</math> which is generated by the revolution of the region bounded by {{nowrap begin}}<math>y=r-\frac{r}{h}x</math>{{nowrap end}} and the lines {{nowrap begin}}<math>y=0</math>{{nowrap end}} and {{nowrap begin}}<math>x=0</math>{{nowrap end}} around the <math>x</math>-axis.|answer=The region extends in the <math>x</math>-direction from {{nowrap begin}}<math>x=0</math>{{nowrap end}} to {{nowrap begin}}<math>x=h</math>{{nowrap end}}. The volume of the solid of revolution is given by
:<math>\begin{align}\int_{0}^{h}\pi\left(r-\frac{r}{h}x\right)^{2}dx&=\pi\int_{0}^{h}\left(r^{2}-\frac{2r^{2}}{h}x+\frac{r^{2}}{h^{2}}x^{2}\right)dx\\
&=\pi r^{2}\left(\int_{0}^{h}dx-\frac{2}{h}\int_{0}^{h}xdx+\frac{1}{h^{2}}\int_{0}^{h}x^{2}dx\right)\\
&=\pi r^{2}\left(x\bigr|_{0}^{h}-\frac{x^{2}}{h}\Bigr|_{0}^{h}+\frac{x^{3}}{3h^{2}}\Bigr|_{0}^{h}\right)\\
&=\pi r^{2}\left(h-h+\frac{h}{3}\right)\\
&=\mathbf{\frac{\pi r^{2}h}{3}}\end{align}</math>}}

{{question-answer|question=2. Calculate the volume of the solid of revolution generated by revolving the region bounded by the curve {{nowrap begin}}<math>y=x^2</math>{{nowrap end}} and the lines {{nowrap begin}}<math>x=1</math>{{nowrap end}} and {{nowrap begin}}<math>y=0</math>{{nowrap end}} around the <math>x</math>-axis.|answer=The region extends in the <math>x</math>-direction from {{nowrap begin}}<math>x=0</math>{{nowrap end}} to {{nowrap begin}}<math>x=1</math>{{nowrap end}}. The volume of the solid of revolution is given by
:<math>\begin{align}\int_{0}^{1}\pi\left(x^{2}\right)^{2}dx&=\pi\int_{0}^{1}x^{4}dx\\
&=\pi\frac{x^{5}}{5}\biggr|_{0}^{1}\\
&=\mathbf{\frac{\pi}{5}}\end{align}</math>}}

{{question-answer|question=3. Use the washer method to find the volume of a cone containing a central hole formed by revolving the region bounded by {{nowrap begin}}<math>y=R-\frac{R}{h}x</math>{{nowrap end}} and the lines {{nowrap begin}}<math>y=r</math>{{nowrap end}} and {{nowrap begin}}<math>x=0</math>{{nowrap end}} around the <math>x</math>-axis.|answer=The <math>x</math> values of the region extend from {{nowrap begin}}<math>x=0</math>{{nowrap end}} to {{nowrap begin}}<math>x=h</math>{{nowrap end}}.  The volume is
:<math>\begin{align}V&=\int_{0}^{h}\pi\left(\left(R-\frac{R}{h}x\right)^{2}-r^{2}\right)dx\\
&=\int_{0}^{h}\pi\left(R^{2}-2\frac{R^{2}}{h}x+\frac{R^{2}}{h^{2}}x^{2}-r^{2}\right)dx\\
&=\pi\left(R^{2}x-\frac{R^{2}}{h}x^{2}+\frac{R^{2}}{3h^{2}}x^{3}-r^{2}x\right)\biggr|_{0}^{h}\\
&=\pi\left(R^{2}h-R^{2}h+\frac{R^{2}h}{3}-r^{2}h\right)\\
&=\mathbf{\pi h\left(\frac{R^{2}}{3}-r^{2}\right)}\end{align}</math>}}

{{question-answer|question=4. Calculate the volume of the solid of revolution generated by revolving the region bounded by the curves {{nowrap begin}}<math>y=x^2</math>{{nowrap end}} and {{nowrap begin}}<math>y=x^3</math>{{nowrap end}} and the lines {{nowrap begin}}<math>x=1</math>{{nowrap end}} and {{nowrap begin}}<math>y=0</math>{{nowrap end}} around the <math>x</math>-axis.|answer=
:<math>\begin{align}V&=\int_{0}^{1}\pi\left(\left(x^{2}\right){}^{2}-\left(x^{3}\right){}^{2}\right)dx\\
&=\int_{0}^{1}\pi\left(x^{4}-x^{6}\right)dx\\
&=\pi\left(\frac{x^{5}}{5}-\frac{x^{7}}{7}\right)\biggr|_{0}^{1}\\
&=\pi\left(\frac{1}{5}-\frac{1}{7}\right)\\
&=\pi\frac{7-5}{35}\\
&=\mathbf{\frac{2\pi}{35}}\end{align}</math>}}

{{question-answer|question=5. Find the volume of a cone with radius <math>R</math> and height <math>h</math> by using the shell method on the appropriate region which, when rotated around the <math>y</math>-axis, produces a cone with the given characteristics.|answer=You could set up the appropriate region in any of the four quadrants. Here we set it up in the first quadrant. Since we are revolving around the <math>y</math>-axis, the <math>y</math> direction will be the height and the radius will be along the <math>x</math> direction. So we need a line that passes through the points {{nowrap begin}}<math>(0,h)</math>{{nowrap end}} and {{nowrap begin}}<math>(R,0)</math>{{nowrap end}}. The slope of this line is
:<math>m=\frac{0-h}{R-0}=-\frac{h}{R}</math>
and the <math>y</math>-intercept is <math>h</math>. Thus, the equation of the line is
:<math>y=-\frac{h}{R}x+h</math>
The <math>x</math>-values of the region run from {{nowrap begin}}<math>x=0</math>{{nowrap end}} to {{nowrap begin}}<math>x=R</math>{{nowrap end}}. Since the function is positive throughout the region we can drop the absolute value sign. The volume will be
:<math>\begin{align}V&=\int_{0}^{R}2\pi x(-\frac{h}{R}x+h)dx\\
&=\int_{0}^{R}2\pi(-\frac{h}{R}x^{2}+hx)dx\\
&=2\pi(-\frac{h}{3R}x^{3}+\frac{h}{2}x^{2})\biggr|_{0}^{R}\\
&=2\pi(-\frac{hR^{2}}{3}+\frac{hR^{2}}{2})\\
&=2\pi\frac{-2hR^{2}+3hR^{2}}{6}\\
&=\mathbf{\frac{\pi hR^{2}}{3}}\end{align}</math>}}

{{question-answer|question=6. Calculate the volume of the solid of revolution generated by revolving the region bounded by the curve {{nowrap begin}}<math>y=x^2</math>{{nowrap end}} and the lines {{nowrap begin}}<math>x=1</math>{{nowrap end}} and {{nowrap begin}}<math>y=0</math>{{nowrap end}} around the <math>y</math>-axis.|answer=
:<math>\begin{align}V&=\int_{0}^{1}2\pi xx^{2}dx\\
&=\int_{0}^{1}2\pi x^{3}dx\\
&=2\pi\frac{x^{4}}{4}\biggr|_{0}^{1}\\
&=\mathbf{\frac{\pi}{2}}\end{align}</math>}}

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