Volume of Region around x=1

[karush]

Find the volume of a region bounded by $y=\sqrt{x}$, $y=1$ and the $y-axis$ that revoles around $x=1$

$$x_1=y^2$$

$$\int_{0}^{1} \left(x_1-1\right)^2\,dy =\frac{8\pi}{15}$$

Book answer is $$\frac{7\pi}{15}$$ ?

 

[MarkFL]

If we choose the shell method here, with:

\(\displaystyle dV=2\pi rh\,dx\)

\(\displaystyle r=1-x\)

\(\displaystyle h=1-y=1-x^{\frac{1}{2}}\)

And so, we have:

\(\displaystyle V=2\pi\int_0^1 (1-x)\left(1-x^{\frac{1}{2}}\right)\,dx=2\pi\int_0^1 1-x^{\frac{1}{2}}-x+x^{\frac{3}{2}}\,dx=2\pi\left[x-\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{2}x^2+\frac{2}{5}x^{\frac{5}{2}}\right]_0^1=2\pi\left(1-\frac{2}{3}-\frac{1}{2}+\frac{2}{5}\right)=\frac{\pi}{15}\left(30-20-15+12\right)=\frac{7\pi}{15}\)

If we choose the washer method instead, we then have:

\(\displaystyle dV=\pi\left(R^2-r^2\right)\,dy\)

\(\displaystyle R=1\)

\(\displaystyle r=1-x=1-y^2\)

And so we have:

\(\displaystyle V=\pi\int_0^1 1^2-(1-y^2)^2\,dy=\pi\int_0^1 2y^2-y^4\,dy=\pi\left[\frac{2}{3}y^3-\frac{1}{5}y^5\right]_0^1=\pi\left(\frac{2}{3}-\frac{1}{5}\right)=\frac{\pi}{15}(10-3)=\frac{7\pi}{15}\)

 

[karush]

Thanks just wasn't a good example in the book
So that really helps a lot mahalo

 

[suluclac]

We could also use the disc (washer) method.
\(\displaystyle \pi(\int_0^1dy-\int_0^1(1-y^2)^2dy)=\pi(1-\int_0^1(y^4-2y^2+1)\,dy)=\pi(1-[\frac{1}{5}y^5-\frac{2}{3}y^3+y\Big|_0^1)=\pi(1-\frac{1}{5}+\frac{2}{3}-1)=\pi(\frac{2}{3}-\frac{1}{5})\)
\(\displaystyle \pi(\frac{2}{3}-\frac{1}{5})=\frac{7\pi}{15}\)

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If we choose the washer method instead, we then have:

\(\displaystyle dV=\pi\left(R^2-r^2\right)\,dy\)

\(\displaystyle R=1\)

\(\displaystyle r=1-x=1-y^2\)

And so we have:

\(\displaystyle V=\pi\int_0^1 1^2-(1-y^2)^2\,dy=\pi\int_0^1 2y^2-y^4\,dy=\pi\left[\frac{2}{3}y^3-\frac{1}{5}y^5\right]_0^1=\pi\left(\frac{2}{3}-\frac{1}{5}\right)=\frac{\pi}{15}(10-3)=\frac{7\pi}{15}\)

But I see that Mark has already posted the correct washer method.