Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. y2=x,x=2yy 2 =x,x=2y; about the y-axis

Answer 1

Answer:

Step-by-step explanation:

From the first image attached below, we will see the sketch of the curve x = y² & x = 2y

In the picture connected underneath, the concealed locale(shaded region) is bounded by the given curves. Now, we discover the marks of the crossing point of the curves. These curves will cross, when:

$y^2 =2 y \\ \\ or y^2 -2y = 0 \\ \\ y(y-2) = 0 \\ \\ y = 0 \ \ or \ \ y = 2$

Thus, the shaded region fall within the interval 0 ≤ y ≤ 2

Now, from the subsequent picture appended we sketch the solid acquired by turning the concealed region about the y-axis.

For the cross-sectional area of the washer:

$A (y) = \pi (outer \ radius)^2 - \pi ( inner \ radius )^2 \\ \\ A(y) = \pi (2y^)2- \pi (y^)^2 \\ \\ A(y) = 4 \pi y^2 - \pi y^4 \\ \\ A(y) = \pi( 4 y^2 -y^4)$

Finally, the volume of (solid) is:

$V = \int^2_0 A(y) \ dy \\ \\ V = \int^2_0 \pi (4y^2 -y^4) \ dy \\ \\ V = \pi \int^2_0 (4y^2 -y^4) \ dy \\ \\ V = \pi \Big[\dfrac{4}{3}y^3 - \dfrac{y^5}{5} \big ] ^2_0 \\ \\ V = \pi \Big [ \dfrac{4}{3}(2)^3-\dfrac{2^3}{5} \Big ] \\ \\ V = \dfrac{64}{15}\pi \\ \\ V = (4.27 ) \pi$