Question

find the volume of the solid generated by revolving the following region about the given axis. the region in the first quadrant bounded above by the curve ( y = x^2 ), below by the x-axis, and on the right by the line ( x = 2 ), about the line ( x = -4 ). ( v = square ) (type an exact answer, using ( pi ) as needed.)

Explanation:

Step1: Identify the method (Washer or Cylindrical Shells)

We can use the method of cylindrical shells or the washer method. Let's use the washer method (or the method of disks/washers with horizontal slices, but since we have a vertical axis of revolution \(x = - 4\) and the region is bounded by \(x\) from \(0\) to \(2\), the shell method might be more straightforward. Wait, actually, for the shell method, the formula for the volume when rotating about a vertical line \(x = a\) is \(V=\int_{c}^{d}2\pi(\text{radius})(\text{height})dx\). The radius of each shell is the distance from \(x\) to \(x=-4\), which is \(x - (-4)=x + 4\). The height of each shell is the function \(y=x^{2}\) (since it's bounded above by \(y = x^{2}\) and below by \(y = 0\) in the first quadrant). The limits of integration are from \(x = 0\) to \(x=2\) (since it's bounded on the right by \(x = 2\) and in the first quadrant, \(x\geq0\)).

Step2: Set up the integral for the shell method

The formula for the volume using the cylindrical shell method when rotating about \(x=-4\) is:
\(V=\int_{0}^{2}2\pi(x + 4)(x^{2})dx\)
First, expand the integrand: \((x + 4)x^{2}=x^{3}+4x^{2}\)
So the integral becomes \(V = 2\pi\int_{0}^{2}(x^{3}+4x^{2})dx\)

Step3: Integrate the function

Integrate term by term. The integral of \(x^{n}\) is \(\frac{x^{n + 1}}{n+1}\) (\(n
eq - 1\)).
\(\int(x^{3}+4x^{2})dx=\frac{x^{4}}{4}+4\times\frac{x^{3}}{3}+C=\frac{x^{4}}{4}+\frac{4x^{3}}{3}+C\)

Step4: Evaluate the definite integral from 0 to 2

Evaluate at the upper limit \(x = 2\):
\(\frac{(2)^{4}}{4}+\frac{4\times(2)^{3}}{3}=\frac{16}{4}+\frac{4\times8}{3}=4+\frac{32}{3}=\frac{12 + 32}{3}=\frac{44}{3}\)
Evaluate at the lower limit \(x = 0\):
\(\frac{0^{4}}{4}+\frac{4\times0^{3}}{3}=0\)
Subtract the lower limit result from the upper limit result: \(\frac{44}{3}-0=\frac{44}{3}\)

Step5: Multiply by \(2\pi\)

\(V=2\pi\times\frac{44}{3}=\frac{88\pi}{3}\)

Wait, let's check using the washer method. If we use the washer method, we need to express \(x\) in terms of \(y\). The curve \(y=x^{2}\) implies \(x=\sqrt{y}\) (since \(x\geq0\) in the first quadrant). The outer radius \(R(y)\) is the distance from \(x=-4\) to \(x = 2\), which is \(2-(-4)=6\). The inner radius \(r(y)\) is the distance from \(x=-4\) to \(x=\sqrt{y}\), which is \(\sqrt{y}-(-4)=\sqrt{y}+4\). The limits of \(y\) are from \(y = 0\) (since below the \(x\)-axis is \(y = 0\)) to \(y=(2)^{2}=4\) (since at \(x = 2\), \(y = 4\)). The volume formula for the washer method is \(V=\pi\int_{a}^{b}(R(y)^{2}-r(y)^{2})dy\)
So \(R(y)=6\), \(r(y)=\sqrt{y}+4\)
\(R(y)^{2}-r(y)^{2}=36-(\sqrt{y}+4)^{2}=36-(y + 8\sqrt{y}+16)=20 - y-8\sqrt{y}\)
Then \(V=\pi\int_{0}^{4}(20 - y-8y^{\frac{1}{2}})dy\)
Integrate term by term:
\(\int(20 - y-8y^{\frac{1}{2}})dy=20y-\frac{y^{2}}{2}-8\times\frac{2}{3}y^{\frac{3}{2}}+C=20y-\frac{y^{2}}{2}-\frac{16}{3}y^{\frac{3}{2}}+C\)
Evaluate from \(0\) to \(4\):
At \(y = 4\): \(20\times4-\frac{4^{2}}{2}-\frac{16}{3}\times4^{\frac{3}{2}}=80 - 8-\frac{16}{3}\times8=72-\frac{128}{3}=\frac{216 - 128}{3}=\frac{88}{3}\)
At \(y = 0\): \(0\)
So \(V=\pi\times\frac{88}{3}=\frac{88\pi}{3}\), which matches the shell method result.

Answer:

\(\dfrac{88\pi}{3}\)