Math / Calculus

QuestionThe answer above is NOT correct.
Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^{6}, y=1$ , and the $y$ -axis and whose cross-sections perpendicular to the $x$ axis are squares.

Studdy Solution

STEP 1

1. The base of the solid is in the first quadrant.
2. The region is bounded by $y = x^6$ , $y = 1$ , and the $y$ -axis.
3. Cross-sections perpendicular to the $x$ -axis are squares.

STEP 2

1. Determine the bounds for integration.
2. Express the side length of the square in terms of $x$ .
3. Set up the integral for the volume.
4. Evaluate the integral.

STEP 3

Determine the bounds for integration:
The region is bounded by $y = x^6$ , $y = 1$ , and the $y$ -axis. Since $y = 1$ is the upper bound and $y = x^6$ is the lower bound, we find the intersection of $y = x^6$ and $y = 1$ :
$x^6 = 1$ $x = 1$
Thus, the bounds for $x$ are from $0$ to $1$ .

STEP 4

Express the side length of the square in terms of $x$ :
The side length of the square is determined by the distance between $y = x^6$ and $y = 1$ . Since each cross-section is a square, the side length $s$ is:
$s = 1 - x^6$

STEP 5

Set up the integral for the volume:
The area of each square cross-section is $s^2$ . Therefore, the volume $V$ is given by:
$V = \int_{0}^{1} (1 - x^6)^2 \, dx$

STEP 6

Evaluate the integral:
First, expand the integrand:
$(1 - x^6)^2 = 1 - 2x^6 + x^{12}$
Now, integrate term by term:
$V = \int_{0}^{1} (1 - 2x^6 + x^{12}) \, dx$
$= \left[ x - \frac{2}{7}x^7 + \frac{1}{13}x^{13} \right]_{0}^{1}$
Evaluate the definite integral:
$V = \left( 1 - \frac{2}{7} \times 1 + \frac{1}{13} \times 1 \right) - \left( 0 - 0 + 0 \right)$
$= 1 - \frac{2}{7} + \frac{1}{13}$
Combine the fractions:
$= \frac{91}{91} - \frac{26}{91} + \frac{7}{91}$
$= \frac{91 - 26 + 7}{91}$
$= \frac{72}{91}$
The volume of the solid is:
$\boxed{\frac{72}{91}}$

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QuestionThe answer above is NOT correct.
Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^{6}, y=1$ , and the $y$ -axis and whose cross-sections perpendicular to the $x$ axis are squares.

Studdy Solution

STEP 1

1. The base of the solid is in the first quadrant.
2. The region is bounded by $y = x^6$ , $y = 1$ , and the $y$ -axis.
3. Cross-sections perpendicular to the $x$ -axis are squares.

STEP 2

1. Determine the bounds for integration.
2. Express the side length of the square in terms of $x$ .
3. Set up the integral for the volume.
4. Evaluate the integral.

STEP 3

Determine the bounds for integration:
The region is bounded by $y = x^6$ , $y = 1$ , and the $y$ -axis. Since $y = 1$ is the upper bound and $y = x^6$ is the lower bound, we find the intersection of $y = x^6$ and $y = 1$ :
$x^6 = 1$ $x = 1$
Thus, the bounds for $x$ are from $0$ to $1$ .

STEP 4

Express the side length of the square in terms of $x$ :
The side length of the square is determined by the distance between $y = x^6$ and $y = 1$ . Since each cross-section is a square, the side length $s$ is:
$s = 1 - x^6$

STEP 5

Set up the integral for the volume:
The area of each square cross-section is $s^2$ . Therefore, the volume $V$ is given by:
$V = \int_{0}^{1} (1 - x^6)^2 \, dx$

STEP 6

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QuestionThe answer above is NOT correct.Find the volume of the solid whose base is the region in the first quadrant bounded by y=x6,y=1y=x^{6}, y=1y=x6,y=1, and the yyy-axis and whose cross-sections perpendicular to the xxx axis are squares.

Studdy Solution

STEP 1

1. The base of the solid is in the first quadrant.2. The region is bounded by y=x6 y = x^6 y=x6, y=1 y = 1 y=1, and the y y y-axis.3. Cross-sections perpendicular to the x x x-axis are squares.

STEP 2

1. Determine the bounds for integration.2. Express the side length of the square in terms of x x x.3. Set up the integral for the volume.4. Evaluate the integral.

STEP 3

STEP 4

Express the side length of the square in terms of x x x:The side length of the square is determined by the distance between y=x6 y = x^6 y=x6 and y=1 y = 1 y=1. Since each cross-section is a square, the side length s s s is:s=1−x6 s = 1 - x^6 s=1−x6

STEP 5

Set up the integral for the volume:The area of each square cross-section is s2 s^2 s2. Therefore, the volume V V V is given by:V=∫01(1−x6)2 dx V = \int_{0}^{1} (1 - x^6)^2 \, dx V=∫01​(1−x6)2dx

STEP 6

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionThe answer above is NOT correct.
Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^{6}, y=1$ , and the $y$ -axis and whose cross-sections perpendicular to the $x$ axis are squares.

1. The base of the solid is in the first quadrant.
2. The region is bounded by $y = x^6$ , $y = 1$ , and the $y$ -axis.
3. Cross-sections perpendicular to the $x$ -axis are squares.

1. Determine the bounds for integration.
2. Express the side length of the square in terms of $x$ .
3. Set up the integral for the volume.
4. Evaluate the integral.

Express the side length of the square in terms of $x$ :
The side length of the square is determined by the distance between $y = x^6$ and $y = 1$ . Since each cross-section is a square, the side length $s$ is:
$s = 1 - x^6$

Set up the integral for the volume:
The area of each square cross-section is $s^2$ . Therefore, the volume $V$ is given by:
$V = \int_{0}^{1} (1 - x^6)^2 \, dx$