Answer to Question #289731 in Calculus for Prof

Question

Determine the volume of the solid/ring obtained by the region bounded by

𝑦=2√𝑥−1and 𝑦=𝑥−1 about line x= -1 using shell method

Answer 1

We have that;

$\displaystyle V=\int^b_aA(y)\ dy, \text{ where }A(y)=\pi((\text{outer radius})^2-(\text{inner radius})^2)$

Now,

$\displaystyle y=2\sqrt{x-1}\Rightarrow x=\frac{y^2}{4}+1\ \text{and }y=x-1\Rightarrow x=y+1\\$

So,

$\displaystyle \frac{y^2}{4}+1=y+1\Rightarrow\frac{y^2}{4}-y=0\Rightarrow y^2-4y=0\Rightarrow y(y-4)=0\Rightarrow y=0,4$

Hence, the first ring will occur at $\displaystyle y=0$ and the final ring will occur at $\displaystyle y=4$ and so these will be our limits of integration.

Next,

outer radius $\displaystyle =y+1+1=y+2$, and inner radius $\displaystyle =\frac{y^2}{4}+1+1=\frac{y^2}{4}+2$

Also, the cross-sectional area is;

$\displaystyle A(y)=\pi((y+2)^2-(\frac{y^2}{4}+2)^2)=\pi\left(4y-\frac{y^4}{16}\right)$

Thus, the volume is;

$\displaystyle V=\int^b_aA(y)\ dy=\pi\int^4_0\left(4y-\frac{y^4}{16}\right)\ dy=\pi\left[2y^2-\frac{y^5}{80}\right]^4_0=\frac{96\pi}{5}$