Question

6.3 homework
question
completed: 5 of 10 | my score: 5/10 pts (50%)
find the volume of the solid generated by revolving the region bounded by the given line and curve about the x - axis.
y = \sqrt{81 - x^{2}}, y = 0
the volume of the solid is
(type an exact answer.)

Explanation:

Step1: Recall volume - of - revolution formula

The disk method formula for the volume $V$ of the solid generated by revolving the region bounded by $y = f(x)$ and $y = 0$ about the $x$-axis from $x=a$ to $x = b$ is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. First, find the $x$-intercepts of $y=\sqrt{81 - x^{2}}$. Set $y = 0$, then $0=\sqrt{81 - x^{2}}$, which gives $x^{2}=81$, so $x=- 9$ and $x = 9$. Here, $f(x)=\sqrt{81 - x^{2}}$.

Step2: Calculate the integral

$V=\pi\int_{-9}^{9}(\sqrt{81 - x^{2}})^{2}dx=\pi\int_{-9}^{9}(81 - x^{2})dx$. Since $\int_{-9}^{9}(81 - x^{2})dx=\int_{-9}^{9}81dx-\int_{-9}^{9}x^{2}dx$. The integral of a constant $\int_{-9}^{9}81dx=81x\big|_{-9}^{9}=81(9-( - 9))=81\times18 = 1458$. The integral of $x^{2}$ is $\int x^{2}dx=\frac{1}{3}x^{3}+C$, so $\int_{-9}^{9}x^{2}dx=\frac{1}{3}x^{3}\big|_{-9}^{9}=\frac{1}{3}(9^{3}-(-9)^{3})=\frac{1}{3}(729 + 729)=\frac{1}{3}\times1458 = 486$. Then $\int_{-9}^{9}(81 - x^{2})dx=1458-486 = 972$.

Step3: Find the volume

$V=\pi\times972 = 972\pi$.

Answer:

$972\pi$