1. Having Trouble with Rotation

Find the volume of the solid that results when the region above the x-axis and below the curve (a>0,b>0) is revolved about the x-axis.

Not really sure about where to start here, any help would be appreciated.[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]

2. Originally Posted by PD1337
Find the volume of the solid that results when the region above the x-axis and below the curve (a>0,b>0) is revolved about the x-axis.
So you want to know the volume that results from rotating that ellipse around the x-axis. First, solve for y, this gives you that $\displaystyle y=\pm \frac{b}{a}\sqrt{a^2-x^2}$.
You'll need only the $\displaystyle +$ case. Given that, the volume of rotation is

$\displaystyle V = 2\pi \cdot \int_0^a \left(\frac{b}{a}\sqrt{a^2-x^2}\right)^2\, dx$
I have taken $\displaystyle 2\pi\cdot\int_0^a\ldots\, dx$ instead of $\displaystyle \pi\cdot\int_{-a}^a\ldots\,dx$ just to slightly simplify the evaluation of the resulting anti-derivative at the lower limit of the integral.
Now you just evaluate that integral and get

$\displaystyle V=2\pi\int_0^a\frac{b^2}{a^2}(a^2-x^2)\, dx=\ldots = \frac{4\pi ab^2}{3}$

Not really sure about where to start here, any help would be appreciated.[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]
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