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The problem goes like this

Rotate the indicated area around the given axis to calculate the volume of the solid of revolution $y=x^2+1$, $x=0$, $x=2$, $y=0$ , around the axis $ y = 5$.

My question is if the solution is given by the following integral (should I take the limits of integration from $0$ to $2$?)

$$2\pi\int_0^2 ((5-y)\sqrt{y-1})\,dy$$

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  • $\begingroup$ Since the rotation is about the horizontal line $y=5$, the integral should be in terms of $dx$. $\endgroup$ – FoiledIt24 Jan 16 at 19:34
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As you are integrating with respect to $y$, it should be $1 \leq y \leq 5$. But you are finding volume of region $2$ rotated around $y = 5$. You can do that but then you need to subtract it from the volume of the cylinder of radius $5$ to find volume of region $1$ that the question asks.

An easier way to do this is to directly find the volume of region $1$ as below -

$4-x^2 \leq r \leq 5$ (distance from $y = 5 \ $ - $ \ $ i) to $y = 0$ which is $5$ and ii) to parabola $y = x^2 + 1$ which is $(5 - (x^2+1) = 4 - x^2$).

Also, $0 \leq x \leq 2$

So the integral is $ \displaystyle \int_0^{2\pi} \int_0^2 \int_{4 - x^2}^5 r \ dr \ dx \ d\theta = \frac{494 \pi}{15}$

Now coming to the integral that you have come up with is based on

$r = 5 - y, 0 \leq x \leq \sqrt{y-1}$

$V = \displaystyle \iiint r \ dx \ d\theta \ dr = \displaystyle \int_1^5 \int_0^{2\pi} \int_0^{\sqrt{y-1}} (5-y) \ dx \ d\theta \ dy$

$V = \displaystyle 2\pi \int_1^5 (5-y) \sqrt{y-1} \ dy = \frac{256 \pi}{15}$

Subtract it from the cylinder volume which is $\displaystyle \int_0^{2\pi} \int_0^5 \int_0^2 y \ dx \ dy \ d\theta = 50 \pi$

So volume of region $1 \displaystyle = 50 \pi - \frac{256 \pi}{15} = \frac{494 \pi}{15}$

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Translate the parabola and the $x$-axis $5$ units down

The area to rotate is limited by

$y=x^2-4,\;y=-5;\;y=0$

Apply the formula here $$V=\pi \int_0^2 \left((-5)^2-\left(x^2-4\right)^2\right) \, dx=\frac{494 \pi }{15}$$

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  • $\begingroup$ +1 Nice answer. You are always very clear in your answers, I really like them! $\endgroup$ – A-Level Student Jan 16 at 19:53
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Be careful that you are calculating "volume", not "area".

The volume is given by $$ \int_0^2 A(x)\,dx $$

where $$ A(x)=\pi[5^2-(5-f(x))^2]^2 $$ with $f(x)=x^2+1$.

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There are (at least) two ways you can go about doing this, and in either case I would make sure you start with drawing a picture.

If we use the washer method we can think of think of splitting the region to be rotated into vertical slabs between $x = 0$ and $x = 2$, where the inner radius is $5 - x^2 - 1$ and the outer radius is $5$. This gives us $$V = \pi\int_{0}^{2}\left(5^2 - (5 - x^2 - 1)^2\right)\,dx =\frac{494\pi}{15}.$$ If we use cylindrical shells, as it seems you are trying to do, we can think of splitting the region rotated into horizontal slabs between $y = 0$ and $y = 5$. If we pick a slab at $y$ it will be a radius of $5 - y$ from the axis of rotation (as you said). To get the height of each slab we need to think about splitting the region into two pieces. Between $y = 0$ and $y=1$ the height of the slab is a constant $2$. Between $y = 1$ and $y = 5$ the height of the slab is the distance between the parabola and the line $x = 2$, which is $2 - \sqrt{y-1}$. So, sing this method, we get $$V = 2\pi\int_{0}^{1}(5-y)(2)\,dy + 2\pi\int_{1}^{5}(5-y)(2-\sqrt{y-1})\,dy = \frac{494\pi}{15}.$$

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