Math / Calculus

Question(1 point) The region bounded by $y=e^{-x^{2}}, y=0, x=0$ , and $x=1$ is revolved about the $y$ -axis. Find the volume of the resulting solid. Answer: $\square$

Studdy Solution

STEP 1

1. We are given the function $y = e^{-x^2}$ .
2. The region is bounded by $y = 0$ , $x = 0$ , and $x = 1$ .
3. The region is revolved about the $y$ -axis.
4. We need to find the volume of the resulting solid.

STEP 2

1. Set up the integral for the volume using the method of cylindrical shells.
2. Evaluate the integral to find the volume.

STEP 3

The method of cylindrical shells is used when revolving around the $y$ -axis. The formula for the volume $V$ is:
$V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$
For this problem, $f(x) = e^{-x^2}$ , $a = 0$ , and $b = 1$ .

STEP 4

Substitute the given function and limits into the formula:
$V = 2\pi \int_{0}^{1} x \cdot e^{-x^2} \, dx$

STEP 5

To evaluate the integral, use substitution. Let $u = -x^2$ , then $du = -2x \, dx$ or $-\frac{1}{2} du = x \, dx$ .
Change the limits of integration: - When $x = 0$ , $u = 0$ . - When $x = 1$ , $u = -1$ .
The integral becomes:
$V = 2\pi \int_{0}^{-1} e^{u} \left(-\frac{1}{2}\right) \, du$
$V = -\pi \int_{0}^{-1} e^{u} \, du$

STEP 6

Evaluate the integral:
$V = -\pi \left[ e^{u} \right]_{0}^{-1}$
$V = -\pi \left( e^{-1} - e^{0} \right)$
$V = -\pi \left( \frac{1}{e} - 1 \right)$
$V = -\pi \left( \frac{1}{e} - \frac{e}{e} \right)$
$V = -\pi \left( \frac{1 - e}{e} \right)$
$V = \pi \left( \frac{e - 1}{e} \right)$
$V = \pi \left( 1 - \frac{1}{e} \right)$
The volume of the resulting solid is:
$\boxed{\pi \left( 1 - \frac{1}{e} \right)}$

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Question(1 point) The region bounded by $y=e^{-x^{2}}, y=0, x=0$ , and $x=1$ is revolved about the $y$ -axis. Find the volume of the resulting solid. Answer: $\square$

Studdy Solution

STEP 1

1. We are given the function $y = e^{-x^2}$ .
2. The region is bounded by $y = 0$ , $x = 0$ , and $x = 1$ .
3. The region is revolved about the $y$ -axis.
4. We need to find the volume of the resulting solid.

STEP 2

1. Set up the integral for the volume using the method of cylindrical shells.
2. Evaluate the integral to find the volume.

STEP 3

The method of cylindrical shells is used when revolving around the $y$ -axis. The formula for the volume $V$ is:
$V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$
For this problem, $f(x) = e^{-x^2}$ , $a = 0$ , and $b = 1$ .

STEP 4

Substitute the given function and limits into the formula:
$V = 2\pi \int_{0}^{1} x \cdot e^{-x^2} \, dx$

STEP 5

STEP 6

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Question(1 point) The region bounded by y=e−x2,y=0,x=0y=e^{-x^{2}}, y=0, x=0y=e−x2,y=0,x=0, and x=1x=1x=1 is revolved about the yyy-axis. Find the volume of the resulting solid. Answer: □\square□

Studdy Solution

STEP 1

1. We are given the function y=e−x2 y = e^{-x^2} y=e−x2.2. The region is bounded by y=0 y = 0 y=0, x=0 x = 0 x=0, and x=1 x = 1 x=1.3. The region is revolved about the y y y-axis.4. We need to find the volume of the resulting solid.

STEP 2

1. Set up the integral for the volume using the method of cylindrical shells.2. Evaluate the integral to find the volume.

STEP 3

The method of cylindrical shells is used when revolving around the y y y-axis. The formula for the volume V V V is:V=2π∫abx⋅f(x) dx V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx V=2π∫ab​x⋅f(x)dxFor this problem, f(x)=e−x2 f(x) = e^{-x^2} f(x)=e−x2, a=0 a = 0 a=0, and b=1 b = 1 b=1.

STEP 4

Substitute the given function and limits into the formula:V=2π∫01x⋅e−x2 dx V = 2\pi \int_{0}^{1} x \cdot e^{-x^2} \, dx V=2π∫01​x⋅e−x2dx

STEP 5

STEP 6

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Question(1 point) The region bounded by $y=e^{-x^{2}}, y=0, x=0$ , and $x=1$ is revolved about the $y$ -axis. Find the volume of the resulting solid. Answer: $\square$

1. We are given the function $y = e^{-x^2}$ .
2. The region is bounded by $y = 0$ , $x = 0$ , and $x = 1$ .
3. The region is revolved about the $y$ -axis.
4. We need to find the volume of the resulting solid.

1. Set up the integral for the volume using the method of cylindrical shells.
2. Evaluate the integral to find the volume.

The method of cylindrical shells is used when revolving around the $y$ -axis. The formula for the volume $V$ is:
$V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$
For this problem, $f(x) = e^{-x^2}$ , $a = 0$ , and $b = 1$ .

Substitute the given function and limits into the formula:
$V = 2\pi \int_{0}^{1} x \cdot e^{-x^2} \, dx$