Questions: Use the region R that is bounded by the graphs of y=1+sqrt(x), x=4, and y=1 to complete the exercises. Region R is revolved about the x-axis to form a solid of revolution whose cross sections are washers. a. What is the outer radius of a cross section of the solid at a point x in [0,4] ? b. What is the inner radius of a cross section of the solid at a point x in [0,4] ? c. What is the area A(x) of a cross section of the solid at a point x in [0,4] ? d. Write an integral for the volume of the solid.

Use the region R that is bounded by the graphs of y=1+sqrt(x), x=4, and y=1 to complete the exercises.
Region R is revolved about the x-axis to form a solid of revolution whose cross sections are washers.
a. What is the outer radius of a cross section of the solid at a point x in [0,4] ?
b. What is the inner radius of a cross section of the solid at a point x in [0,4] ?
c. What is the area A(x) of a cross section of the solid at a point x in [0,4] ?
d. Write an integral for the volume of the solid.

Transcript text: 7-10. Use the region $R$ that is bounded by the graphs of $y=1+\sqrt{x}$, $x=4$, and $y=1$ to complete the exercises. 7. Region $R$ is revolved about the $x$-axis to form a solid of revolution whose cross sections are washers. a. What is the outer radius of a cross section of the solid at a point $x$ in $[0,4]$ ? b. What is the inner radius of a cross section of the solid at a point $x$ in $[0,4]$ ? c. What is the area $A(x)$ of a cross section of the solid at a point $x$ in $[0,4]$ ? d. Write an integral for the volume of the solid.

Solution

Solution Steps

Step 1: Outer Radius

The outer radius is the distance from the $x$-axis ($y=0$) to the curve $y = 1 + \sqrt{x}$. Therefore, the outer radius is $1 + \sqrt{x}$.

Step 2: Inner Radius

The inner radius is the distance from the $x$-axis ($y=0$) to the line $y=1$. Therefore, the inner radius is $1$.

Step 3: Area of Cross Section

The area of a cross section (washer) is given by $A(x) = \pi (\text{outer radius})^2 - \pi (\text{inner radius})^2$. In this case, $A(x) = \pi(1+\sqrt{x})^2 - \pi(1)^2 = \pi[(1 + 2\sqrt{x} + x) - 1] = \pi(2\sqrt{x} + x)$.

Final Answer

a. $1 + \sqrt{x}$ b. $1$ c. $\pi(x + 2\sqrt{x})$

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