Answer to Question #148521 in Calculus for qwerty

Question

Identify the curves (lines, parabola, etc.) generated by illustrating the graphs.

5. Consider the region enclosed by the curves y=√x, y=6-x, and the x-axis. Rotate this region about the x-axis and find the resulting volume.
6. Consider the region enclosed by the curves y=√x, y=6-x, and the x-axis. Rotate this region about the y-axis and find the resulting volume.
7. Consider the region enclosed by the curves y = x^2 − 2x and y = 3. Rotate this region about the line y = 3 and find the resulting volume.
8. Consider the region enclosed by the curves x = y^2 and x = 2 − y^2. Rotate this region about the line x = 3 and find the resulting volume.

Answer 1

$$Solution$$

5. ) Is this the sum of two integrals or is it difference of two integrals?

First determine where the curves intersect: Obviously $y=\sqrt{x}$ meets the x-axis at $x=0$ and $y=x-2$ meets the x-axis at $x=2$. Now $y=x-2$ and $y=\sqrt{x}$ intersect when

$$x-2=\sqrt{x} \implies x^2-4x+4=x \implies x^2-3x+4=(x-4)(x+1)=0\\ \implies x=4, (not\ x=-1)$$

From the sketch above

Left: The region enclosed by the curves $y=\sqrt{x},y=2-x,$ and the x-axis. When rotated about the x-axis, one region must be subtracted from the other. Instead of using representative rectangles, we simply indicate the appropriate radii with arrows.

From the sketch above sketch, if revolved about the x-axis will result in a solid that has been partially hollowed out (a cone has been removed). This requires a difference of integrals.

$V=Outer\ volume-Inner \ volume\\ \ = \pi \intop_0^4[\sqrt x]^2dx-\pi \intop_2^4[x-2]^2dx\\ \ = \pi \intop_0^4 xdx-\pi \intop_2^4[x-2]^2dx\\ \ =\frac{\pi x^2}{2}|^4_0-\frac{\pi(x-2)^3}{3}|^4_2\\ \ = (8\pi - 0)+(\frac{8 \pi}{3}-0)\\ \ = \frac{16 \pi}{3}$

6 ) Rotate the region about they-axis and find the resulting volume.

Is this the sum of two integrals or is it difference of two integrals?

Since the rotation is about the y-axis, the radii of the respective regions are horizontal, see the sketch below.

This is again a difference of two integrals.

Translating the curves into functions of $y$ we have $x=y^2$, $x=y+2$, and $y=0$ (the x-axis). The curves intersect the x-axis at $y=0$. We’ve seen that the line and square root function meet when $x=4$ since there, then they-coordinate of the intersection is $y=2$.

$V=Outer\ volume-Inner\ volume=\pi \intop_0^2[y+2]^2dy-\pi \intop_0^2[y^2]^2dy\\ \ = \frac{\pi(y+2)^3}{3}|_0^2-\frac{\pi y^5}{5}|_0^2\\ \ = (\frac{64 \pi}{3}-\frac{8 \pi}{3})-(\frac{32 \pi}{3}-0)\\ \ = \frac{184 \pi}{15}$

7 ) The two curves meet when

$$x^2-2x=3 \implies x^2-2x-3=(x-3)(x+1)=0\implies x=3,-1$$

The parabola and the line are easy to sketch; see the below sketch on the left

Left: The region enclosed by the curves $y=x^2-2x$ and $y=3$. Since the axis of revolution is $y=3$, a representative radius extends from the line $y=3$ to the curve $y=x^2-2x$. Right: The resulting solid of revolution about the line $y=3.$

A representative radius extends from the line $y=3$ to the curve $y=x^2-2x.$ The length of a radius is the difference between these two values, that is, the radius of a circular cross-section perpendicular to the line $y=3$ is

$$r=3-(x^2-2x)=3+2x-x^2$$

Since a cross-section is a circle, its area is

$$A=A(x)=\pi r^2=\pi(3+2x-x^2)^2$$

Since we know the cross-sectional area, we find the volume

$V=\intop_a^b A(x)dx=\pi \intop_{-1}^3(3+2x-x^2)dx\\ \ = \pi \intop_{-1}^3(9+12x-2x^2-4x^3+x^4)dx\\ \ = \pi (9x+6x^2-\frac{2x^3}{3}-x^4+\frac{x^5}{5})|_{-1}^3\\\\ \ = \pi[(27+54-18-81+\frac{243}{5})-(-9+6+\frac23-1-\frac15)]\\ \ = \frac{512 \pi}{15}$

8. Consider the region enclosed by the curves $x = y^2$ and $x = 2 − y^2.$

The two curves meet when

$$y^2=2-y^2 \implies 2y^2=2 \implies y=\pm 1$$

An ‘outer’ representative radius extends from the line $x=3$ to the curve $x=y^2$ and an ‘inner’ representative radius extends from the line $x=3$ to $x=2-y^2$. The length of a representative radius is the difference between the corresponding pairs of values

Left: The region enclosed by the curves $x=y^2$ and $x=2-y^2$ and two representative radii emanating from the axis of rotation $x=3.$ Right:The resulting ‘hollowed out’ solid of revolution about the line $x=3$ looks like a ‘doughnut.’

The outer radius is $R=3-y^2$ and the inner radius is $3-(2-y^2)=1+y^2.$ The integration will take place along the y-axis on the interval $[-1, 1]$ because the disks are horizontal when points are rotated about the line $x=3.$ Since we know the cross-sections are circles, we can find the volume

$V=Outer\ volume-Inner\ volume=\pi \intop_{-1}^1[3-y^2]^2dy-\pi \intop_{-1}^1[1+y^2]^2dy\\ \ =\pi \intop_{-1}^19-6y^2+y^4\ dy-\pi \intop_{-1}^11+2y^2+y^4\ dy\\ \ =\pi \intop_{-1}^1 8-8y^2 dy\\ \ =\pi(8y-\frac{8y^3}{3})|_{-1}^1\\ \ = [(8- \frac83)-(-8+\frac83)]\\ \ = \frac{32 \pi}{3}$