Questions: Exercise. The region R lies in the first quadrant and is bounded by the curves y=x, y=-x^2+6 and x=0. A solid is formed by revolving R about the line x=4. To set up an integral or sum of integrals with respect to x that would give the volume of the solid, which method should be used to find the volume? - Shell Method - Washer Method Correct How many integrals with respect to x will we need to express the volume of the solid? 1 Exercise. The integral that gives the volume of the solid is: To set up an integral or sum of integrals with respect to y that would give the volume of the solid, which method should be used to find the volume? - Shell Method

Solution Steps

The region $R$ is bounded by $y=x$, $y=-x^2+6$, and $x=0$. To find the intersection points, we set $x = -x^2 + 6$. This gives $x^2 + x - 6 = 0$, which factors as $(x+3)(x-2)=0$. Since we are in the first quadrant, we take $x=2$. The intersection point is $(2,2)$.

Since we are revolving around the vertical line $x=4$, and we are integrating with respect to $x$, the shell method is appropriate. Since the region is defined by a single function of x for the interval, we only need one integral.

The shell method formula is $V = 2\pi \int_a^b r(x)h(x) dx$. Here, $a=0$ and $b=2$. The radius $r(x)$ is the distance from the axis of rotation $x=4$ to the $x$-value, which is $4-x$. The height $h(x)$ is the difference between the two curves, $(-x^2+6) - x$. Thus, the integral is $$ V = 2\pi \int_0^2 (4-x)(-x^2+6-x) dx = 2\pi \int_0^2 (x^3 -5x^2 + 10x +24) dx $$ $$ V = 2\pi\left[\frac{1}{4} x^4 - \frac{5}{3} x^3 + 5x^2 + 24x \right]_0^2 = 2\pi \left(4 - \frac{40}{3} + 20 + 48 \right) = 2\pi \left(72 - \frac{40}{3}\right) = 2\pi \left( \frac{216 - 40}{3} \right) = \frac{352\pi}{3} $$

Since the axis of rotation $x=4$ is vertical, and we are integrating with respect to $y$, we use the washer method. We have two regions to consider: $0 \le y \le 2$ and $2 \le y \le 6$. For $0 \le y \le 2$, $x=y$, so the outer radius is $R = 4-0 = 4$ and the inner radius is $r=4-y$. For $2 \le y \le 6$, $y=-x^2+6 \implies x^2=6-y \implies x = \sqrt{6-y}$. The outer radius is $R=4-0=4$ and the inner radius is $r = 4 - \sqrt{6-y}$. So, the integral is $$ V = \pi \int_0^2 (4^2 - (4-y)^2)dy + \pi \int_2^6 (4^2 - (4-\sqrt{6-y})^2) dy $$ $$ V = \pi \int_0^2 (8y-y^2) dy + \pi \int_2^6 (16 - (16 - 8\sqrt{6-y} + 6-y)) dy $$ $$ V = \pi \int_0^2 (8y-y^2) dy + \pi \int_2^6 (y+8\sqrt{6-y}-6) dy $$

Final Answer