Author Topic: Volume between two curves? (Read 768 times) Tweet Share

Yoda · « **on:** May 15, 2014, 06:11:23 pm »

Need help for the following question
Find the volume for the region enclosed by the graphs y=2x^2 and x^2+y^2/9=1 for y>=0 when the solid is rotated around the y-axis.

Zealous · « **Reply #1 on:** May 16, 2014, 11:32:17 pm »

Quote from: Yoda on May 15, 2014, 06:11:23 pm

Need help for the following question
Find the volume for the region enclosed by the graphs y=2x^2 and x^2+y^2/9=1 for y>=0 when the solid is rotated around the y-axis.

If you look at the graph below you'll see that we can express the volume as the sum of: area A rotated about the y- axis and area B rotated about the y-axis.

You can use your own method to find out that the intersection points of the two graphs are:

$(\frac{\sqrt{3}}{2},\frac{3}{2}) \ , \ (\frac{-\sqrt{3}}{2},\frac{3}{2}) \ , \ (0,3)$

Take note of the first intersection point, because that is most important one!

To find the volume of A rotated about y, first express x in terms of y then revolve it around the y-axis:

$\\\begin{eqnarray*}\\x^{2}+\frac{y^{2}}{9} & = & 1\\x & = & \sqrt{1-\frac{y^{2}}{9}}\\\\V_{A} & = & \pi\int_{\frac{3}{2}}^{3}(\sqrt{1-\frac{y^{2}}{9}})^{2}dy=\frac{\pi(4\pi-3\sqrt{3})}{8}\\\end{eqnarray*}$

Above, we used the terminals 3 to 3/2 as these are the y-values in which we want to rotate about the y-axis. Have a look at the integrals and back to the diagram if you're confused!

Do the same for the volume of B:

$\\\begin{eqnarray*}y & = & 2x^{2}\\x & = & \sqrt{\frac{y}{2}}\\\\V_{B} & = & \pi\int_{0}^{\frac{3}{2}}(\sqrt{\frac{y}{2}})^{2}dy=\frac{9\pi}{16}\\\\V_{A}+V_{B} & = & \frac{\pi(8\pi-6\sqrt{3}+9)}{16}\\\end{eqnarray*}\\\\$

Hopefully that help and there's no errors (I'm pretty tired =o).

Search the forums now!

We have moved!

Author Topic: Volume between two curves? (Read 768 times) Tweet Share

Yoda

Volume between two curves?

Zealous

Re: Volume between two curves?

Recent Posts

User:
Password: