Questions: As a hardworking student, plagued by too much homework, you spend all night doing math homework. By 6am, you imagine yourself to be a region bounded by y=5 x^2 x=0 x=1 y=0 As you grow more and more tired, the world begins to spin around you. However, according to Newton, there is no difference between the world spinning around you, and you spinning around the world. Unfortunately, you are so tired that you think the world is the x-axis. What is the volume of the solid you (the region) create by spinning about the x-axis? Answer exactly or round to 2 decimal places.

As a hardworking student, plagued by too much homework, you spend all night doing math homework. By 6am, you imagine yourself to be a region bounded by

y=5 x^2
x=0
x=1
y=0

As you grow more and more tired, the world begins to spin around you. However, according to Newton, there is no difference between the world spinning around you, and you spinning around the world. Unfortunately, you are so tired that you think the world is the x-axis. What is the volume of the solid you (the region) create by spinning about the x-axis? Answer exactly or round to 2 decimal places.

Transcript text: As a hardworking student, plagued by too much homework, you spend all night doing math homework. By 6am, you imagine yourself to be a region bounded by \[ \begin{array}{l} y=5 x^{2} \\ x=0 \\ x=1 \\ y=0 \end{array} \] As you grow more and more tired, the world begins to spin around you. However, according to Newton, there is no difference between the world spinning around you, and you spinning around the world. Unfortunately, you are so tired that you think the world is the $x$-axis. What is the volume of the solid you (the region) create by spinning about the x-axis? Answer exactly or round to 2 decimal places.

Solution

Solution Steps

Step 1: Set up the integral for the volume of the solid of revolution

Using the disk method, the volume $V$ of the solid formed by rotating the region about the $x$-axis is given by the integral $V = \pi \int_{0}^{1} (5x^2)^2 dx$.

Step 2: Compute the integral

Simplify the integral to $\pi \int_{0}^{1} 5^2x^4 dx$, and then find the antiderivative, which is $\frac{\pi 5^2x^5}{5}$.

Step 3: Evaluate the definite integral

Substitute the limits of integration to find $V = \pi \left[\frac{5^2x^5}{5}\right]_{0}^{1} = \frac{\pi 5^21^5}{5}$.