# Solid of revolution axis $y=5$

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 $$2$$?)

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

• Since the rotation is about the horizontal line $y=5$, the integral should be in terms of $dx$. – FoiledIt24 Jan 16 at 19:34

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 $$1 \displaystyle = 50 \pi - \frac{256 \pi}{15} = \frac{494 \pi}{15}$$

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

• +1 Nice answer. You are always very clear in your answers, I really like them! – A-Level Student Jan 16 at 19:53

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 $$f(x)=x^2+1$$.

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 $$V = 2\pi\int_{0}^{1}(5-y)(2)\,dy + 2\pi\int_{1}^{5}(5-y)(2-\sqrt{y-1})\,dy = \frac{494\pi}{15}.$$