Question

Math

Posted 6 months ago

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Let $R$ be the region enclosed by $y=3 \sqrt{4-x}$ and the axes in the first quadrant.
Region $R$ is the base of a solid. For each $y$-value, the cross section of the solid taken perpendicular to the $y$-axis is a rectangle whose base lies in $R$ and whose height is $y$.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) $\int_{0}^{4}\left(4-\frac{y^{2}}{9}\right) \cdot y d y$
(B) $\int_{0}^{6}\left(4-\frac{y^{2}}{9}\right) \cdot y d y$
(c) $\int_{0}^{4} 3 \sqrt{4-y} \cdot y d y$
() $\int_{0}^{6} 3 \sqrt{4-y} \cdot y d y$
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Answer from Sia

Posted 6 months ago

Solution by Steps

step 2

The cross-section of the solid perpendicular to the y-axis is a rectangle with base in $R$ and height $y$

step 3

The base of the rectangle at a given $y$ is the horizontal distance from the y-axis to the curve, which is $x = 4 - \frac{y^2}{9}$

step 4

The volume of the solid is given by the integral of the area of the cross-section along the y-axis. The area of the cross-section is $y \cdot \text{base} = y \cdot (4 - \frac{y^2}{9})$

step 5

The correct integral to find the volume is $\int_{0}^{6} 3\sqrt{4-y} \cdot y \, dy$

D

Key Concept

Volume of a solid with cross-sectional area

Explanation

The volume of a solid can be found by integrating the area of its cross-sections along the axis perpendicular to the cross-sections.

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