§6.2 Using definite integrals to find volume (continued)

Revolving about horizontal and vertical lines other than the coordinate axes

Summary

The region $S$ between $y = 2x$ and $y = x^3$ in the first quadrant is revolved about the line $y = -2$. A typical washer has

The same region $S$ is revolved about the line $y = 4$. The correct washer radii are

The same region $S$ is revolved about the line $x = -1$. The washer method then uses

When the region $S$ is revolved about the line $x = 5$, the outer radius is $5-\frac{1}{2}y$ and the inner radius is $5-y^{1/3}$, giving volume $\frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677$.

With the washer setup about $y = -2$, the exact volume is $\frac{4}{21}(21+8\sqrt{2})\pi$. Report the decimal, rounded to three places.

The region between $y = x^2$ and $y = x$ is revolved about the line $y = -1$. The correct washer setup and volume are

For $f(x) = 3\cos(\frac{x^3}{4})$ with region $R$ under the first arch on $[0, 1.84527]$, which pairing of quantity and integral is correct?

Using technology on $\int_0^{1.84527}3\cos(\frac{x^3}{4})\,dx$, the area of the region under the first arch of $f(x) = 3\cos(\frac{x^3}{4})$ is about $4.6623$. Report the value, rounded to three decimals.

The region $R$ bounded by the $y$-axis with $y = \sin(x)$ and $y = \cos(x)$ up to their first positive intersection. Revolving about the $x$-axis gives

The region bounded by $y = 1+\frac{1}{2}(x-2)^2$, $y = \frac{1}{2}x^2$, and $x = 0$ is revolved about the line $y = -1$, using washers with outer radius $2+\frac{1}{2}(x-2)^2$ and inner radius $1+\frac{1}{2}x^2$. The exact volume is $\frac{315}{32}\pi$. Report the decimal, rounded to three places.

The curves $y = 1+\frac{1}{2}(x-2)^2$ and $y = \frac{1}{2}x^2$ meet at $x = \frac{3}{2}$. Compute the area of the region they bound with the $y$-axis, given $A = \int_0^{1.5}\left(1+\frac{1}{2}(x-2)^2-\frac{1}{2}x^2\right)dx$.