Determine the volumes of the solids generated by revolving the regions
bounded by the specified curves about the indicated axis of rotation y-axis
a. A region enclosed by a triangle with vertices and

Ask by Powers Marshall. in South Africa

Mar 23,2025

Question

Determine the volumes of the solids generated by revolving the regions
bounded by the specified curves about the indicated axis of rotation y-axis
a. A region enclosed by a triangle with vertices and

Ask by Powers Marshall. in South Africa

Mar 23,2025

UpStudy AI · Accepted Answer

We begin by sketching the triangle with vertices $$ (1,1) $$, $$ (2,1) $$, and $$ (1,0) $$. The sides of the triangle are:

- The horizontal segment from $$ (1,1) $$ to $$ (2,1) $$ (where $$ y=1 $$),
- The vertical segment from $$ (1,1) $$ to $$ (1,0) $$ (where $$ x=1 $$),
- The slanted segment from $$ (1,0) $$ to $$ (2,1) $$.

The line joining $$ (1,0) $$ and $$ (2,1) $$ has slope
$$
\frac{1-0}{2-1}=1,
$$
so its equation is
$$
y=x-1.
$$

Since we are revolving the region about the $$ y $$-axis, it is convenient to use the shell method. Notice that for a fixed $$ x $$ in the interval $$ [1,2] $$, the vertical extent of the region is from the line $$ y=x-1 $$ (the lower boundary) up to the horizontal line $$ y=1 $$ (the upper boundary). The height of the shell is therefore
$$
\text{height} = 1 - (x-1) = 2 - x.
$$

A typical shell at $$ x $$ has:
- Radius: $$ x $$,
- Height: $$ 2-x $$,
- Thickness: $$ dx $$.

The formula for the volume using the shell method is
$$
V = 2\pi \int_{a}^{b} (\text{radius})(\text{height})\,dx.
$$
Substitute the expressions and the limits $$ x=1 $$ to $$ x=2 $$:
$$
V = 2\pi \int_{1}^{2} x\,(2-x)\,dx.
$$

Expand the integrand:
$$
x(2-x)=2x-x^2.
$$
Thus,
$$
V = 2\pi \int_{1}^{2} (2x - x^2)\,dx.
$$

Now, compute the integral:
$$
\int (2x - x^2)\,dx = x^2 - \frac{x^3}{3} + C.
$$
Evaluate from $$ x=1 $$ to $$ x=2 $$:
$$
\left[ x^2 - \frac{x^3}{3} \right]_{1}^{2} 
= \left( 2^2 - \frac{2^3}{3} \right) - \left( 1^2 - \frac{1^3}{3} \right) 
= \left( 4 - \frac{8}{3} \right) - \left( 1 - \frac{1}{3} \right).
$$
Write each term with a common denominator:
$$
4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3},
$$
$$
1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}.
$$
Thus, the definite integral is
$$
\frac{4}{3} - \frac{2}{3} = \frac{2}{3}.
$$

Substitute back into the volume formula:
$$
V = 2\pi \cdot \frac{2}{3} = \frac{4\pi}{3}.
$$

Thus, the volume of the solid obtained by revolving the given triangle about the $$ y $$-axis is
$$
\boxed{\frac{4\pi}{3}}.
$$
The volume of the solid generated by revolving the triangle about the y-axis is $$ \frac{4\pi}{3} $$.

Determine the volumes of the solids generated by revolving the regions
bounded by the specified curves about the indicated axis of rotation y-axis
a. A region enclosed by a triangle with vertices and

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Determine the volumes of the solids generated by revolving the regions bounded by the specified curves about the indicated axis of rotation y-axis a. A region enclosed by a triangle with vertices and

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Determine the volumes of the solids generated by revolving the regions
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a. A region enclosed by a triangle with vertices and