Volume of Region Bounded by $y = \sqrt{x}$ and $y = 18-x^2$

Question

I haven't a clue why I'm having so much issues solving this problem. The two curves are $y=\sqrt{x}$ and $y=18-x^2, x \ge 1$, and rotated about the $y$-axis.

If I'm not mistaken, I need to use the shell method. In that case, the volume integration formula is $\int 2\pi x f(x)dx$.

So my thickness is $dx$, my height is $f(x)$, I believe my range is from $1$ to $\sqrt{18}$ ($1$ because $x \ge 1$ and $\sqrt{18}$ because $0 = 18-x^2$).

So I integrate $2\pi \cdot x \cdot x^{1/2} dx$ and get $(2/5)x^{5/2}$ from $1$ to $\sqrt{18}$, but that isn't even close to the answer.

I've been working on this problem far too long and am getting brain fog.

Answer 1

You will be adding "a lot of" silinders of height $dy$ with radius $r(y)$. It is of the form $$V = \int _0^2 S_1(y)dy + \int _2^{18} S_2(y)dy $$ The two curves meet at $y=2$ (for $x=4$) so for $y\in [0,2)$ the radius of the circle is determined by $y=\sqrt{x}$, therefore $r_1(y) = y^2, y\in [0,2)$. The radius is determined by $y=18-x^2$ for $y\in [2,18]$ hence $r_2(y) = \sqrt{18-y}, y\in [2,18]$.

Our volume is therefore $$\pi\int_0^2 y^4dy + \pi\int_2^{18} (18-y)dy\tag{1} $$

If $x\geq 1$ is assumed, then we need to subtract some volume. To picture this, draw a vertical axis at $x=1$ and subtract the volume which is obtained by revolution of the piece to the left of $x=1$. We compute the volume similarly so obtain $$\pi\int_0^1 y^4dy + \pi\int_1^{17} dy + \pi\int_{17}^{18}(18-y)dy \tag{2}$$ Volume of interest is obtained by subtracting $(2)$ from $(1)$.

Answer 2

Note that the graph of $y = \sqrt{x}$ and $y = 18-x^2$ shows that these two curves meet at $x = 4$ and $y = 18-4^2 = \sqrt{4} = 2$. Write out explicitly $f$ in the question body.

$$f(x) = 18-x^2-\sqrt{x} \quad \forall\,x \in [1,4]$$

\begin{align} \text{Required volume of revolution} &= 2\pi \int_1^4 x f(x) dx \\ &= 2\pi \int_1^4 (18x - x^{3/2} - x^3) dx \\ &= 2\pi \left[9x^2 - \frac{x^{5/2}}{5/2} - \frac{x^4}{4} \right]^4_1 \\ &= 2\pi \left[ 9(4^2 - 1^2) - \frac{64}{5} + \frac25 - \frac{4^4 - 1^4}{4} \right] \\ &= \frac{1177\pi}{10} \end{align}

Answer 3

Look at the region and set up the integral as below:

$$V = 2\pi\int_{1}^{4} (18-x^2-\sqrt{x})xdx$$

$$V = 2\pi \int_1^4 x (18-x^2) dx - 2\pi \int_1^4 x \sqrt{x} dx $$ $$== 2\pi \left[ 9(4^2-1) - \frac{4^4-1}{4} -\frac{64}{5} +\frac25 \right]$$

$$ V = \frac{1177\pi}{10}$$