I don't know how to find the volume enclosed by this surface of revolution, obtained by revolving the region bounded by the curves $$\begin{cases} x = 5\cos t \\ y=3\sin t \end{cases},\ x = 0,\ x = 2$$ about the line $ x=2$
What integral should I use? Help, please.
On
The parametric equation that you have is the equation of an ellipse. Just eliminate $t$. $$(x/5)^2=\cos^2t\\(y/3)^2=\sin^2t\\\left(\frac{x}{5}\right)^2+\left(\frac{y}{3}\right)^2=1$$ We can easily calculate the intersections of the ellipse with $x=2$, and you get two solutions, one positive, one negative: $y_{int}=\pm\frac{3}{5}\sqrt{21}$.
When you rotate your figure, you get a cylinder with some "dents" at the top and bottom. Each of these dents can be understood as made out of disks with radius $2-|x(y)|$ and thickness $dy$. So the volume is $$V=\pi2^26-2V_{dent}$$ where $$V_{dent}=\int_{y_{int}}^3\pi(2-|x(y)|)^2dy$$ Just in case you did not guess $$x(y)=5\sqrt{1-y^2/9}$$
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Disks/washers or shells?
Lets go with shells.
$A = 2\pi \int xy dx\\ x = 5\cos t\\ dx = -5\sin t \ dt\\ y = 3\sin t$
Limits $\arccos (\frac 25)$ to $\pi$
This is only half way around so I will need to double it.
$4\pi\displaystyle \int_{\arccos (\frac 25)}^{\pi} -75\sin^2 t\cos t dt\\ 4\pi (-25\sin^3 t)|_{\arccos (\frac 25)}^{\pi}\\ 4\pi (25)(1-\frac {4}{25})^{\frac 32})\\ 4\pi (\frac {21 \sqrt {21}}{5})$
First, rewrite the ellipse equation
$$ \frac{x^2}{25} + \frac{y^2}{9} = 1$$
or, in terms of $x$ $$ y = \pm 3 \sqrt{1 - \frac{x^2}{25}} $$
Using the cylindrical shell method, we add up shells with thichness $dx$, radius $r=2-x$ and height $h = 2|y| = 6\sqrt{1-\dfrac{x^2}{25}}$. The total volume is then
$$ V = \int 2\pi r h\ dx = 2\pi\int_0^2 6(2-x)\sqrt{1-\frac{x^2}{25}}\ dx $$