QuestionFind the volume $V$ of the solid formed by revolving the area under $y=10 e^{-x}$ in the first quadrant around the $y$ -axis.

Studdy Solution

STEP 1

Assumptions1. The curve is given by $y=10e^{-x}$ . . The region under the curve in the first quadrant is revolved around the $y$ -axis.
3. We are looking for the volume $V$ of the solid generated by this revolution.

STEP 2

We will use the method of cylindrical shells to find the volume. The formula for the volume of a solid of revolution using cylindrical shells is given by $V =2\pi \int_a^b x \cdot f(x) \, dx$ where $f(x)$ is the function defining the curve, and $a$ and $b$ are the limits of the region being revolved.

STEP 3

In our case, the function $f(x)$ is given by $10e^{-x}$ . The region under the curve in the first quadrant is from $x=0$ to $x=\infty$ . So, our limits of integration are $a=0$ and $b=\infty$ .

STEP 4

Substitute $f(x)$ , $a$ , and $b$ into the volume formula $V =2\pi \int0^\infty x \cdot10e^{-x} \, dx$

STEP 5

implify the integral $V =20\pi \int0^\infty x e^{-x} \, dx$

STEP 6

This integral is not straightforward to evaluate, so we will use integration by parts. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$ where $u$ and $v$ are functions of $x$ . We choose $u=x$ and $dv=e^{-x} \, dx$ .

STEP 7

Compute $du$ and $v$ $du = dx, \quad v = -e^{-x}$

STEP 8

Substitute $u$ , $v$ , $du$ , and $dv$ into the integration by parts formula $\int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx$

STEP 9

implify the integral $\int x e^{-x} \, dx = -x e^{-x} + \int e^{-x} \, dx$

STEP 10

Evaluate the remaining integral $\int e^{-x} \, dx = -e^{-x}$

STEP 11

Substitute this result back into the equation from9 $\int x e^{-x} \, dx = -x e^{-x} - (-e^{-x}) = -x e^{-x} + e^{-x}$

STEP 12

Now we can substitute this result back into our volume formula from5 $V =20\pi \left[ -x e^{-x} + e^{-x} \right]0^\infty$

STEP 13

Evaluate the limits of the integral $V =20\pi \left[ \lim_{x \to \infty} (-x e^{-x} + e^{-x}) - \lim_{x \to0} (-x e^{-x} + e^{-x}) \right]$

STEP 14

Evaluate the limits $V =20\pi \left[0 - (0 +) \right] = -20\pi$

STEP 15

The volume of a solid cannot be negative, so we take the absolute value to get the final answer $V = | -20\pi | =20\pi$ So, the volume $V$ of the solid generated by revolving the region under the curve $y=10e^{-x}$ in the first quadrant about the $y$ -axis is $20\pi$ cubic units.

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