Math  /  Calculus

QuestionFind the volume of the solid obtained by rotating the region in the first quadrant bounded by x=0,y=1,x=y15x=0, y=1, x=y^{15} about the line y=10y=10.

Studdy Solution

STEP 1

1. The region is in the first quadrant.
2. The region is bounded by x=0, y=1, and x=y^15.
3. We are rotating this region about the line y=10.
4. We will use the method of cylindrical shells to solve this problem.

STEP 2

1. Visualize and sketch the region.
2. Set up the integral for the volume using the method of cylindrical shells.
3. Determine the limits of integration.
4. Evaluate the integral.
5. Simplify and calculate the final result.

STEP 3

Sketch the region bounded by x=0, y=1, and x=y^15 in the first quadrant. Note that the curve x=y^15 intersects the line y=1 when x=1.

STEP 4

We will use the method of cylindrical shells. The general formula for this method is:
V=2πabr(y)h(y)dy V = 2\pi \int_a^b r(y) h(y) dy
Where: - r(y) is the distance from the axis of rotation to the shell - h(y) is the height of the shell
In our case: - The axis of rotation is y=10 - r(y) = 10 - y (distance from y=10 to the shell) - h(y) = y^15 (the x-coordinate of the boundary curve)
Our integral becomes:
V=2πab(10y)y15dy V = 2\pi \int_a^b (10-y) y^{15} dy

STEP 5

Determine the limits of integration: - The lower limit is y=0 (where the region starts at the x-axis) - The upper limit is y=1 (where the curve intersects the line y=1)
Our integral is now:
V=2π01(10y)y15dy V = 2\pi \int_0^1 (10-y) y^{15} dy

STEP 6

Evaluate the integral:
V=2π01(10y15y16)dy V = 2\pi \int_0^1 (10y^{15} - y^{16}) dy
V=2π[10y1616y1717]01 V = 2\pi \left[ \frac{10y^{16}}{16} - \frac{y^{17}}{17} \right]_0^1

STEP 7

Simplify and calculate the final result:
V=2π[(1016117)(00)] V = 2\pi \left[ \left(\frac{10}{16} - \frac{1}{17}\right) - (0 - 0) \right]
V=2π(1016117) V = 2\pi \left( \frac{10}{16} - \frac{1}{17} \right)
V=2π(17027216272) V = 2\pi \left( \frac{170}{272} - \frac{16}{272} \right)
V=2π(154272) V = 2\pi \left( \frac{154}{272} \right)
V=2π(77136) V = 2\pi \left( \frac{77}{136} \right)
V=77π68 V = \frac{77\pi}{68}
The volume of the solid is:
77π68 \boxed{\frac{77\pi}{68}}

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